Did you know? The tangent to a circle is perpendicular to the radius at the point of contact. You can prove this with implicit differentiation, and you can also use that geometry to check your derivative. On the circle x²+y²=25, at (3,4), the tangent gradient is −3/4 and the normal gradient is 4/3. The normal line passes through the centre (0,0), but the tangent does not.
This JC2 H2 Mathematics Tutor Bukit Timah lesson answers a focused A-Level 9758 question: how do you differentiate an implicit circle equation, write its tangent and normal, and handle vertical tangents without dividing by zero? The quick rule comes first, followed by worked examples, geometric checks and 36 original coordinate clinics. Each problem uses a fictional circle, not a map, surveyed property or actual student result.
The 2027 SEAB H2 Mathematics syllabus 9758 includes implicit differentiation and tangents and normals to curves. This guide builds on that requirement. It also uses the familiar geometry of a circle to verify the result independently. There is no need to find a second derivative of a parametrically defined curve, which the syllabus explicitly excludes. The goal is a correct derivative and an interpretable line equation.
Quick answer: differentiate both x and y terms
For the circle (x−h)²+(y−k)²=r², differentiate implicitly with respect to x. The x term gives 2(x−h); the y term gives 2(y−k)(dy/dx), by the chain rule. Thus dy/dx=−(x−h)/(y−k) wherever y≠k. At a point P=(x₀,y₀) on the circle, the tangent is perpendicular to the radius vector (x₀−h,y₀−k). The normal line points along that radius.
- Confirm the point: substitute P into the original circle equation before differentiating.
- Tangent gradient: mₜ=−(x₀−h)/(y₀−k) when y₀≠k.
- Normal gradient: mₙ=(y₀−k)/(x₀−h) when x₀≠h.
- Exception: when y₀=k, the tangent is vertical, and dy/dx is not finite there.
- Exception: when x₀=h, the tangent is horizontal, and the normal is vertical.
- Robust tangent equation: (x₀−h)(x−x₀)+(y₀−k)(y−y₀)=0; it works in all nondegenerate cases.
- Robust normal: (x,y)=(x₀,y₀)+t(x₀−h,y₀−k), t∈ℝ.
Complete worked example: x²+y²=25 at (3,4)
Substitute (3,4): 3²+4²=9+16=25, so the point is on the circle. Differentiate x²+y²=25 to obtain 2x+2y(dy/dx)=0. Rearranging gives dy/dx=−x/y where y≠0. At P=(3,4), the tangent gradient is −3/4. Through the point (3,4), the tangent equation is y−4=−(3/4)(x−3), or equivalently 3x+4y=25.
The radius from the centre to P has direction vector (3,4), with gradient 4/3. Its line equation is y−4=(4/3)(x−3), or 4x−3y=0. This is the normal. Their gradients multiply to −1, confirming perpendicularity. The tangent is also perpendicular to the radius because the normal is radial. Two independent checks agree with the differentiation.
Why the tangent equation avoids dividing by zero
The dot product of the radius vector (a,b) with any direction along the tangent is zero. If P=(h+a,k+b), this yields the tangent equation a(x−h−a)+b(y−k−b)=0, or a(x−x₀)+b(y−y₀)=0. This formula works whether a or b is zero, as long as (a,b) is nonzero. It is more robust than starting with a slope formula and dividing by a coordinate that might vanish.
At the rightmost point of x²+y²=25, P=(5,0), the radius vector is (5,0) and the tangent equation is 5(x−5)=0, or x=5. This is vertical, and no finite dy/dx exists there. Treating the derivative as zero would give a horizontal line, which is exactly the wrong geometry. The corresponding normal is the horizontal line y=0.
A horizontal tangent and a vertical normal
At the top of x²+y²=25, P=(0,5), the radius points vertically. The derivative dy/dx=−x/y gives zero, so the tangent is the horizontal line y=5. The normal runs vertically through the centre, so its equation is x=0. Its gradient is undefined as a finite number; that is not a mathematical defect of the line, just a reason to use a vertical-line equation instead of y=mx+c.
This pair of extreme points gives a mental picture for every circle. At left and right extremes, the tangent is vertical. At upper and lower extremes, the tangent is horizontal. A student should be able to state these conclusions before differentiating a single symbol. The derivative confirms the geometry.
Shifted centres: the same method, changed coordinates
For a circle centred at (2,−1), consider (x−2)²+(y+1)²=25 and P=(5,3). Its radius vector from the centre is (3,4). Implicit differentiation gives dy/dx=−(x−2)/(y+1), so the tangent gradient at P is −3/4. The tangent is 3(x−5)+4(y−3)=0, or 3x+4y=27. The normal passes through both P and the centre: (x,y)=(5,3)+t(3,4).
The gradient pattern is unchanged because only the centre has moved. A student’s calculation should depend on the offset from the centre, not the raw coordinates of P alone. Substituting 5 and 3 directly into −x/y would give −5/3, incorrect for a shifted circle. The correct formula includes x−h and y−k.
A non-point is not a place for a tangent
Suppose a question says “find the tangent to x²+y²=25 at (4,4)”. The point does not lie on the circle, because 4²+4²=32 rather than 25. Therefore there is no tangent at that point of the circle. A tangent from an external point may be a different problem, possibly with two solutions, but you must not insert the stated off-curve coordinates into the derivative and assert that the resulting line is a tangent at a nonexistent contact point.
This is why point verification belongs before differentiation. It takes seconds and can prevent an entire solution being built on an inconsistent premise. If the question asks for tangent lines from an external point, the required contact points must be found by a different method; the wording “at P” and “through P” are not equivalent.
A tangent is locally straight, even when the curve bends
Implicit differentiation gives a gradient at a point of the curve. The tangent line is the local linear approximation there; it is not the same as the circle itself and does not match the curve at a second arbitrary point. The normal line is perpendicular to this local tangent. On a circle, it happens to pass through the centre because the radius is a normal direction.
That last property is special to circles. For a general implicit curve, a normal to the curve need not pass through any global “centre”. The method of differentiating both x and y terms still works, but the geometry of radial normals is a convenient independent check only in the circle setting. A strong learner distinguishes a general calculus method from a special geometric shortcut.
The one-minute diagnostic
Ask a learner about x²+y²=25 at (3,4): derivative −3/4, tangent 3x+4y=25, normal through the centre with slope 4/3. Next change P to (5,0). The tangent becomes vertical x=5, not horizontal. Finally change the circle centre to (2,−1) and ask what goes into the derivative: (x−2) and (y+1), not x and y. These three changes test method, exceptional points and correct coordinate offsets.
Thirty-six original circle tangent-and-normal clinics
Each problem below supplies a circle by centre and radius and a point with integer offsets from the centre. You can verify its location using the Pythagorean relationship. The tangent and normal are then obtained from the radius offset. Every clinic has a consistent algebraic and geometric check. These are original teaching examples rather than official past-year examination questions.
1. Circle centre (-4,-5), contact (-1,-9)
Question. Consider the fictional circle (x+4)²+(y+5)²=25 with centre (-4,-5). Find the tangent and normal at P=(-1,-9). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=3 and b=-4. Their squares add to 9+16=25, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-4))/(y−(-5)); at P, the tangent gradient is 3/4 and the normal gradient is -4/3. The tangent line is 3(x−(-1))+-4(y−(-9))=0, and the normal line can be written (x,y)=(-1,-9)+t(3,-4).
Geometric audit. The tangent is perpendicular to the centre-to-point radius. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (3,-4). The normal through P has direction (3,-4) and therefore passes through (-4,-5) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
2. Circle centre (-3,-2), contact (1,1)
Question. Consider the fictional circle (x+3)²+(y+2)²=25 with centre (-3,-2). Find the tangent and normal at P=(1,1). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=4 and b=3. Their squares add to 16+9=25, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-3))/(y−(-2)); at P, the tangent gradient is -4/3 and the normal gradient is 3/4. The tangent line is 4(x−(1))+3(y−(1))=0, and the normal line can be written (x,y)=(1,1)+t(4,3).
Geometric audit. The normal line passes through the circle’s centre. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (4,3). The normal through P has direction (4,3) and therefore passes through (-3,-2) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
3. Circle centre (-2,1), contact (-7,13)
Question. Consider the fictional circle (x+2)²+(y−1)²=169 with centre (-2,1). Find the tangent and normal at P=(-7,13). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-5 and b=12. Their squares add to 25+144=169, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-2))/(y−(1)); at P, the tangent gradient is 5/12 and the normal gradient is -12/5. The tangent line is -5(x−(-7))+12(y−(13))=0, and the normal line can be written (x,y)=(-7,13)+t(-5,12).
Geometric audit. Reversing both offset signs gives the opposite point on the circle with parallel tangent. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-5,12). The normal through P has direction (-5,12) and therefore passes through (-2,1) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
4. Circle centre (-1,4), contact (-13,-1)
Question. Consider the fictional circle (x+1)²+(y−4)²=169 with centre (-1,4). Find the tangent and normal at P=(-13,-1). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-12 and b=-5. Their squares add to 144+25=169, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-1))/(y−(4)); at P, the tangent gradient is -12/5 and the normal gradient is 5/12. The tangent line is -12(x−(-13))+-5(y−(-1))=0, and the normal line can be written (x,y)=(-13,-1)+t(-12,-5).
Geometric audit. A translated circle uses offsets from its centre, not raw x and y coordinates. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-12,-5). The normal through P has direction (-12,-5) and therefore passes through (-1,4) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
5. Circle centre (0,-4), contact (8,11)
Question. Consider the fictional circle x²+(y+4)²=289 with centre (0,-4). Find the tangent and normal at P=(8,11). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=8 and b=15. Their squares add to 64+225=289, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(0))/(y−(-4)); at P, the tangent gradient is -8/15 and the normal gradient is 15/8. The tangent line is 8(x−(8))+15(y−(11))=0, and the normal line can be written (x,y)=(8,11)+t(8,15).
Geometric audit. The radius vector dotted with a tangent direction is zero. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (8,15). The normal through P has direction (8,15) and therefore passes through (0,-4) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
6. Circle centre (1,-1), contact (16,7)
Question. Consider the fictional circle (x−1)²+(y+1)²=289 with centre (1,-1). Find the tangent and normal at P=(16,7). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=15 and b=8. Their squares add to 225+64=289, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(1))/(y−(-1)); at P, the tangent gradient is -15/8 and the normal gradient is 8/15. The tangent line is 15(x−(16))+8(y−(7))=0, and the normal line can be written (x,y)=(16,7)+t(15,8).
Geometric audit. A gradient-based solution must handle vertical tangents separately. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (15,8). The normal through P has direction (15,8) and therefore passes through (1,-1) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
7. Circle centre (2,2), contact (-5,-22)
Question. Consider the fictional circle (x−2)²+(y−2)²=625 with centre (2,2). Find the tangent and normal at P=(-5,-22). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-7 and b=-24. Their squares add to 49+576=625, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(2))/(y−(2)); at P, the tangent gradient is -7/24 and the normal gradient is 24/7. The tangent line is -7(x−(-5))+-24(y−(-22))=0, and the normal line can be written (x,y)=(-5,-22)+t(-7,-24).
Geometric audit. The squared radius is a²+b², which can be checked before differentiating. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-7,-24). The normal through P has direction (-7,-24) and therefore passes through (2,2) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
8. Circle centre (3,5), contact (-21,12)
Question. Consider the fictional circle (x−3)²+(y−5)²=625 with centre (3,5). Find the tangent and normal at P=(-21,12). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-24 and b=7. Their squares add to 576+49=625, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(3))/(y−(5)); at P, the tangent gradient is 24/7 and the normal gradient is -7/24. The tangent line is -24(x−(-21))+7(y−(12))=0, and the normal line can be written (x,y)=(-21,12)+t(-24,7).
Geometric audit. The normal’s direction is radial and can be expressed without a slope. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-24,7). The normal through P has direction (-24,7) and therefore passes through (3,5) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
9. Circle centre (4,-3), contact (13,9)
Question. Consider the fictional circle (x−4)²+(y+3)²=225 with centre (4,-3). Find the tangent and normal at P=(13,9). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=9 and b=12. Their squares add to 81+144=225, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(4))/(y−(-3)); at P, the tangent gradient is -3/4 and the normal gradient is 4/3. The tangent line is 9(x−(13))+12(y−(9))=0, and the normal line can be written (x,y)=(13,9)+t(9,12).
Geometric audit. An exact fraction is preferable to an early rounded gradient. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (9,12). The normal through P has direction (9,12) and therefore passes through (4,-3) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
10. Circle centre (-4,0), contact (8,-9)
Question. Consider the fictional circle (x+4)²+y²=225 with centre (-4,0). Find the tangent and normal at P=(8,-9). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=12 and b=-9. Their squares add to 144+81=225, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-4))/(y−(0)); at P, the tangent gradient is 4/3 and the normal gradient is -3/4. The tangent line is 12(x−(8))+-9(y−(-9))=0, and the normal line can be written (x,y)=(8,-9)+t(12,-9).
Geometric audit. The robust dot-product tangent equation works even when one coordinate offset is zero. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (12,-9). The normal through P has direction (12,-9) and therefore passes through (-4,0) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
11. Circle centre (-3,3), contact (-9,11)
Question. Consider the fictional circle (x+3)²+(y−3)²=100 with centre (-3,3). Find the tangent and normal at P=(-9,11). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-6 and b=8. Their squares add to 36+64=100, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-3))/(y−(3)); at P, the tangent gradient is 3/4 and the normal gradient is -4/3. The tangent line is -6(x−(-9))+8(y−(11))=0, and the normal line can be written (x,y)=(-9,11)+t(-6,8).
Geometric audit. The tangent is perpendicular to the centre-to-point radius. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-6,8). The normal through P has direction (-6,8) and therefore passes through (-3,3) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
12. Circle centre (-2,-5), contact (-10,1)
Question. Consider the fictional circle (x+2)²+(y+5)²=100 with centre (-2,-5). Find the tangent and normal at P=(-10,1). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-8 and b=6. Their squares add to 64+36=100, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-2))/(y−(-5)); at P, the tangent gradient is 4/3 and the normal gradient is -3/4. The tangent line is -8(x−(-10))+6(y−(1))=0, and the normal line can be written (x,y)=(-10,1)+t(-8,6).
Geometric audit. The normal line passes through the circle’s centre. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-8,6). The normal through P has direction (-8,6) and therefore passes through (-2,-5) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
13. Circle centre (-1,-2), contact (2,-6)
Question. Consider the fictional circle (x+1)²+(y+2)²=25 with centre (-1,-2). Find the tangent and normal at P=(2,-6). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=3 and b=-4. Their squares add to 9+16=25, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-1))/(y−(-2)); at P, the tangent gradient is 3/4 and the normal gradient is -4/3. The tangent line is 3(x−(2))+-4(y−(-6))=0, and the normal line can be written (x,y)=(2,-6)+t(3,-4).
Geometric audit. Reversing both offset signs gives the opposite point on the circle with parallel tangent. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (3,-4). The normal through P has direction (3,-4) and therefore passes through (-1,-2) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
14. Circle centre (0,1), contact (4,4)
Question. Consider the fictional circle x²+(y−1)²=25 with centre (0,1). Find the tangent and normal at P=(4,4). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=4 and b=3. Their squares add to 16+9=25, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(0))/(y−(1)); at P, the tangent gradient is -4/3 and the normal gradient is 3/4. The tangent line is 4(x−(4))+3(y−(4))=0, and the normal line can be written (x,y)=(4,4)+t(4,3).
Geometric audit. A translated circle uses offsets from its centre, not raw x and y coordinates. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (4,3). The normal through P has direction (4,3) and therefore passes through (0,1) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
15. Circle centre (1,4), contact (-4,16)
Question. Consider the fictional circle (x−1)²+(y−4)²=169 with centre (1,4). Find the tangent and normal at P=(-4,16). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-5 and b=12. Their squares add to 25+144=169, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(1))/(y−(4)); at P, the tangent gradient is 5/12 and the normal gradient is -12/5. The tangent line is -5(x−(-4))+12(y−(16))=0, and the normal line can be written (x,y)=(-4,16)+t(-5,12).
Geometric audit. The radius vector dotted with a tangent direction is zero. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-5,12). The normal through P has direction (-5,12) and therefore passes through (1,4) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
16. Circle centre (2,-4), contact (-10,-9)
Question. Consider the fictional circle (x−2)²+(y+4)²=169 with centre (2,-4). Find the tangent and normal at P=(-10,-9). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-12 and b=-5. Their squares add to 144+25=169, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(2))/(y−(-4)); at P, the tangent gradient is -12/5 and the normal gradient is 5/12. The tangent line is -12(x−(-10))+-5(y−(-9))=0, and the normal line can be written (x,y)=(-10,-9)+t(-12,-5).
Geometric audit. A gradient-based solution must handle vertical tangents separately. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-12,-5). The normal through P has direction (-12,-5) and therefore passes through (2,-4) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
17. Circle centre (3,-1), contact (11,14)
Question. Consider the fictional circle (x−3)²+(y+1)²=289 with centre (3,-1). Find the tangent and normal at P=(11,14). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=8 and b=15. Their squares add to 64+225=289, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(3))/(y−(-1)); at P, the tangent gradient is -8/15 and the normal gradient is 15/8. The tangent line is 8(x−(11))+15(y−(14))=0, and the normal line can be written (x,y)=(11,14)+t(8,15).
Geometric audit. The squared radius is a²+b², which can be checked before differentiating. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (8,15). The normal through P has direction (8,15) and therefore passes through (3,-1) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
18. Circle centre (4,2), contact (19,10)
Question. Consider the fictional circle (x−4)²+(y−2)²=289 with centre (4,2). Find the tangent and normal at P=(19,10). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=15 and b=8. Their squares add to 225+64=289, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(4))/(y−(2)); at P, the tangent gradient is -15/8 and the normal gradient is 8/15. The tangent line is 15(x−(19))+8(y−(10))=0, and the normal line can be written (x,y)=(19,10)+t(15,8).
Geometric audit. The normal’s direction is radial and can be expressed without a slope. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (15,8). The normal through P has direction (15,8) and therefore passes through (4,2) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
19. Circle centre (-4,5), contact (-11,-19)
Question. Consider the fictional circle (x+4)²+(y−5)²=625 with centre (-4,5). Find the tangent and normal at P=(-11,-19). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-7 and b=-24. Their squares add to 49+576=625, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-4))/(y−(5)); at P, the tangent gradient is -7/24 and the normal gradient is 24/7. The tangent line is -7(x−(-11))+-24(y−(-19))=0, and the normal line can be written (x,y)=(-11,-19)+t(-7,-24).
Geometric audit. An exact fraction is preferable to an early rounded gradient. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-7,-24). The normal through P has direction (-7,-24) and therefore passes through (-4,5) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
20. Circle centre (-3,-3), contact (-27,4)
Question. Consider the fictional circle (x+3)²+(y+3)²=625 with centre (-3,-3). Find the tangent and normal at P=(-27,4). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-24 and b=7. Their squares add to 576+49=625, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-3))/(y−(-3)); at P, the tangent gradient is 24/7 and the normal gradient is -7/24. The tangent line is -24(x−(-27))+7(y−(4))=0, and the normal line can be written (x,y)=(-27,4)+t(-24,7).
Geometric audit. The robust dot-product tangent equation works even when one coordinate offset is zero. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-24,7). The normal through P has direction (-24,7) and therefore passes through (-3,-3) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
21. Circle centre (-2,0), contact (7,12)
Question. Consider the fictional circle (x+2)²+y²=225 with centre (-2,0). Find the tangent and normal at P=(7,12). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=9 and b=12. Their squares add to 81+144=225, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-2))/(y−(0)); at P, the tangent gradient is -3/4 and the normal gradient is 4/3. The tangent line is 9(x−(7))+12(y−(12))=0, and the normal line can be written (x,y)=(7,12)+t(9,12).
Geometric audit. The tangent is perpendicular to the centre-to-point radius. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (9,12). The normal through P has direction (9,12) and therefore passes through (-2,0) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
22. Circle centre (-1,3), contact (11,-6)
Question. Consider the fictional circle (x+1)²+(y−3)²=225 with centre (-1,3). Find the tangent and normal at P=(11,-6). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=12 and b=-9. Their squares add to 144+81=225, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-1))/(y−(3)); at P, the tangent gradient is 4/3 and the normal gradient is -3/4. The tangent line is 12(x−(11))+-9(y−(-6))=0, and the normal line can be written (x,y)=(11,-6)+t(12,-9).
Geometric audit. The normal line passes through the circle’s centre. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (12,-9). The normal through P has direction (12,-9) and therefore passes through (-1,3) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
23. Circle centre (0,-5), contact (-6,3)
Question. Consider the fictional circle x²+(y+5)²=100 with centre (0,-5). Find the tangent and normal at P=(-6,3). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-6 and b=8. Their squares add to 36+64=100, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(0))/(y−(-5)); at P, the tangent gradient is 3/4 and the normal gradient is -4/3. The tangent line is -6(x−(-6))+8(y−(3))=0, and the normal line can be written (x,y)=(-6,3)+t(-6,8).
Geometric audit. Reversing both offset signs gives the opposite point on the circle with parallel tangent. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-6,8). The normal through P has direction (-6,8) and therefore passes through (0,-5) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
24. Circle centre (1,-2), contact (-7,4)
Question. Consider the fictional circle (x−1)²+(y+2)²=100 with centre (1,-2). Find the tangent and normal at P=(-7,4). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-8 and b=6. Their squares add to 64+36=100, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(1))/(y−(-2)); at P, the tangent gradient is 4/3 and the normal gradient is -3/4. The tangent line is -8(x−(-7))+6(y−(4))=0, and the normal line can be written (x,y)=(-7,4)+t(-8,6).
Geometric audit. A translated circle uses offsets from its centre, not raw x and y coordinates. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-8,6). The normal through P has direction (-8,6) and therefore passes through (1,-2) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
25. Circle centre (2,1), contact (5,-3)
Question. Consider the fictional circle (x−2)²+(y−1)²=25 with centre (2,1). Find the tangent and normal at P=(5,-3). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=3 and b=-4. Their squares add to 9+16=25, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(2))/(y−(1)); at P, the tangent gradient is 3/4 and the normal gradient is -4/3. The tangent line is 3(x−(5))+-4(y−(-3))=0, and the normal line can be written (x,y)=(5,-3)+t(3,-4).
Geometric audit. The radius vector dotted with a tangent direction is zero. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (3,-4). The normal through P has direction (3,-4) and therefore passes through (2,1) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
26. Circle centre (3,4), contact (7,7)
Question. Consider the fictional circle (x−3)²+(y−4)²=25 with centre (3,4). Find the tangent and normal at P=(7,7). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=4 and b=3. Their squares add to 16+9=25, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(3))/(y−(4)); at P, the tangent gradient is -4/3 and the normal gradient is 3/4. The tangent line is 4(x−(7))+3(y−(7))=0, and the normal line can be written (x,y)=(7,7)+t(4,3).
Geometric audit. A gradient-based solution must handle vertical tangents separately. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (4,3). The normal through P has direction (4,3) and therefore passes through (3,4) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
27. Circle centre (4,-4), contact (-1,8)
Question. Consider the fictional circle (x−4)²+(y+4)²=169 with centre (4,-4). Find the tangent and normal at P=(-1,8). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-5 and b=12. Their squares add to 25+144=169, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(4))/(y−(-4)); at P, the tangent gradient is 5/12 and the normal gradient is -12/5. The tangent line is -5(x−(-1))+12(y−(8))=0, and the normal line can be written (x,y)=(-1,8)+t(-5,12).
Geometric audit. The squared radius is a²+b², which can be checked before differentiating. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-5,12). The normal through P has direction (-5,12) and therefore passes through (4,-4) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
28. Circle centre (-4,-1), contact (-16,-6)
Question. Consider the fictional circle (x+4)²+(y+1)²=169 with centre (-4,-1). Find the tangent and normal at P=(-16,-6). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-12 and b=-5. Their squares add to 144+25=169, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-4))/(y−(-1)); at P, the tangent gradient is -12/5 and the normal gradient is 5/12. The tangent line is -12(x−(-16))+-5(y−(-6))=0, and the normal line can be written (x,y)=(-16,-6)+t(-12,-5).
Geometric audit. The normal’s direction is radial and can be expressed without a slope. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-12,-5). The normal through P has direction (-12,-5) and therefore passes through (-4,-1) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
29. Circle centre (-3,2), contact (5,17)
Question. Consider the fictional circle (x+3)²+(y−2)²=289 with centre (-3,2). Find the tangent and normal at P=(5,17). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=8 and b=15. Their squares add to 64+225=289, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-3))/(y−(2)); at P, the tangent gradient is -8/15 and the normal gradient is 15/8. The tangent line is 8(x−(5))+15(y−(17))=0, and the normal line can be written (x,y)=(5,17)+t(8,15).
Geometric audit. An exact fraction is preferable to an early rounded gradient. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (8,15). The normal through P has direction (8,15) and therefore passes through (-3,2) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
30. Circle centre (-2,5), contact (13,13)
Question. Consider the fictional circle (x+2)²+(y−5)²=289 with centre (-2,5). Find the tangent and normal at P=(13,13). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=15 and b=8. Their squares add to 225+64=289, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-2))/(y−(5)); at P, the tangent gradient is -15/8 and the normal gradient is 8/15. The tangent line is 15(x−(13))+8(y−(13))=0, and the normal line can be written (x,y)=(13,13)+t(15,8).
Geometric audit. The robust dot-product tangent equation works even when one coordinate offset is zero. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (15,8). The normal through P has direction (15,8) and therefore passes through (-2,5) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
31. Circle centre (-1,-3), contact (-8,-27)
Question. Consider the fictional circle (x+1)²+(y+3)²=625 with centre (-1,-3). Find the tangent and normal at P=(-8,-27). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-7 and b=-24. Their squares add to 49+576=625, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(-1))/(y−(-3)); at P, the tangent gradient is -7/24 and the normal gradient is 24/7. The tangent line is -7(x−(-8))+-24(y−(-27))=0, and the normal line can be written (x,y)=(-8,-27)+t(-7,-24).
Geometric audit. The tangent is perpendicular to the centre-to-point radius. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-7,-24). The normal through P has direction (-7,-24) and therefore passes through (-1,-3) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
32. Circle centre (0,0), contact (-24,7)
Question. Consider the fictional circle x²+y²=625 with centre (0,0). Find the tangent and normal at P=(-24,7). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-24 and b=7. Their squares add to 576+49=625, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(0))/(y−(0)); at P, the tangent gradient is 24/7 and the normal gradient is -7/24. The tangent line is -24(x−(-24))+7(y−(7))=0, and the normal line can be written (x,y)=(-24,7)+t(-24,7).
Geometric audit. The normal line passes through the circle’s centre. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-24,7). The normal through P has direction (-24,7) and therefore passes through (0,0) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
33. Circle centre (1,3), contact (10,15)
Question. Consider the fictional circle (x−1)²+(y−3)²=225 with centre (1,3). Find the tangent and normal at P=(10,15). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=9 and b=12. Their squares add to 81+144=225, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(1))/(y−(3)); at P, the tangent gradient is -3/4 and the normal gradient is 4/3. The tangent line is 9(x−(10))+12(y−(15))=0, and the normal line can be written (x,y)=(10,15)+t(9,12).
Geometric audit. Reversing both offset signs gives the opposite point on the circle with parallel tangent. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (9,12). The normal through P has direction (9,12) and therefore passes through (1,3) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
34. Circle centre (2,-5), contact (14,-14)
Question. Consider the fictional circle (x−2)²+(y+5)²=225 with centre (2,-5). Find the tangent and normal at P=(14,-14). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=12 and b=-9. Their squares add to 144+81=225, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(2))/(y−(-5)); at P, the tangent gradient is 4/3 and the normal gradient is -3/4. The tangent line is 12(x−(14))+-9(y−(-14))=0, and the normal line can be written (x,y)=(14,-14)+t(12,-9).
Geometric audit. A translated circle uses offsets from its centre, not raw x and y coordinates. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (12,-9). The normal through P has direction (12,-9) and therefore passes through (2,-5) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
35. Circle centre (3,-2), contact (-3,6)
Question. Consider the fictional circle (x−3)²+(y+2)²=100 with centre (3,-2). Find the tangent and normal at P=(-3,6). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-6 and b=8. Their squares add to 36+64=100, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(3))/(y−(-2)); at P, the tangent gradient is 3/4 and the normal gradient is -4/3. The tangent line is -6(x−(-3))+8(y−(6))=0, and the normal line can be written (x,y)=(-3,6)+t(-6,8).
Geometric audit. The radius vector dotted with a tangent direction is zero. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-6,8). The normal through P has direction (-6,8) and therefore passes through (3,-2) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
36. Circle centre (4,1), contact (-4,7)
Question. Consider the fictional circle (x−4)²+(y−1)²=100 with centre (4,1). Find the tangent and normal at P=(-4,7). Before applying a gradient, verify that the point lies on the circle by comparing the sum of the squared offsets with the stated squared radius. Then choose either an implicit derivative or the normal-vector tangent equation.
Full calculation. The radius offsets are a=-8 and b=6. Their squares add to 64+36=100, so P is on the circle. Implicit differentiation gives dy/dx=−(x−(4))/(y−(1)); at P, the tangent gradient is 4/3 and the normal gradient is -3/4. The tangent line is -8(x−(-4))+6(y−(7))=0, and the normal line can be written (x,y)=(-4,7)+t(-8,6).
Geometric audit. A gradient-based solution must handle vertical tangents separately. The tangent equation uses the radius offset as its normal vector, so a direction along the tangent must be perpendicular to (-8,6). The normal through P has direction (-8,6) and therefore passes through (4,1) when its parameter is chosen appropriately. Changing the sign of the radius offset changes the orientation of this vector, not the validity of the perpendicular relationship.
Four extreme-point cases without finite gradient
- For x²+y²=25 at (5,0): tangent x=5, normal y=0.
- At (−5,0): tangent x=−5, normal y=0.
- At (0,5): tangent y=5, normal x=0.
- At (0,−5): tangent y=−5, normal x=0.
Five-day practice sequence
Day one: verify that proposed points actually lie on their circles. Day two: differentiate unshifted equations x²+y²=r² and write tangent equations from gradients. Day three: shift the circle centre and use x−h, y−k consistently. Day four: practise the four extreme points where a finite tangent or normal gradient fails, using vertical and horizontal line equations. Day five: solve mixed questions without being told whether the tangent is horizontal, vertical or oblique, checking every answer with the radius perpendicularity condition.
In a three-pupil tutorial, one learner can verify the contact point, another can perform implicit differentiation and a third can derive the tangent directly from the radius. The independent methods must agree. Rotate roles and change the centre; this makes a hidden coordinate-offset error visible before it carries into a complete solution.
Frequently asked questions
What is the derivative of x²+y²=r²?
It is 2x+2y(dy/dx)=0, so dy/dx=−x/y where y≠0.
Why is the tangent perpendicular to the radius?
The radius is a normal direction to the circle at its contact point; tangent directions have zero scalar product with it.
How do I find a tangent at a vertical point?
Use the normal-vector equation a(x−x₀)+b(y−y₀)=0, which works when the usual derivative is not finite.
Does the normal always go through the centre?
For a circle, yes. The normal direction is radial from the centre to the contact point.
Must I verify that P is on the circle?
Yes. A line claimed to touch the circle at P requires P to lie on that circle.
Is implicit differentiation part of the 2027 H2 syllabus?
Yes. The syllabus includes differentiation of simple implicit functions and tangents and normals to curves.
Related guides and syllabus
- Composite Functions: Hidden Domain After a Square Root
- Implicit Differentiation: Tangent and Normal to a Circle: current guide
- Without Replacement: Dependent Draws
- Zero Correlation Does Not Prove Independence
For the broader topic, see JC H2 Implicit and Parametric Differentiation, Bukit Timah JC2 Tuition, and the immutable Clementi small-group mathematics tutoring example for the teaching format.
Official source: SEAB H2 Mathematics 9758 syllabus (2027), Differentiation. All circles and coordinates are fictional mathematical exercises, not actual buildings or recorded student outcomes.