A tangent touches a circle at one point and is perpendicular to the radius drawn to that point of contact. Coordinate geometry turns this geometric fact into an efficient method for finding tangent equations.
1. Radius and tangent are perpendicular
For a circle with centre C and tangent point P, line CP is perpendicular to the tangent at P.
If the radius gradient is m and is finite and nonzero, the tangent gradient is −1/m.
2. Worked tangent from a known contact point
Find the tangent to (x−1)²+(y−2)²=25 at P(4,6).
The centre is C(1,2). Gradient CP=(6−2)/(4−1)=4/3. Therefore the tangent gradient is −3/4.
Using point-gradient form, y−6=−(3/4)(x−4). Hence 3x+4y−36=0.
3. Verify the contact point first
Before finding a tangent at a stated point, confirm that the point lies on the circle if this is not guaranteed. For P(4,6), (4−1)²+(6−2)²=9+16=25.
4. Horizontal and vertical radius cases
If the radius to the contact point is horizontal, the tangent is vertical. If the radius is vertical, the tangent is horizontal.
For x²+y²=16 at (4,0), the radius lies along y=0, so the tangent is x=4.
5. Tangent to x²+y²=r²
At a point (x₁,y₁) on x²+y²=r², a compact tangent form is xx₁+yy₁=r².
At (3,4) on x²+y²=25, the tangent is 3x+4y=25.
6. Shifted-circle tangent form
For (x−a)²+(y−b)²=r² and contact point (x₁,y₁), translate to centre-based coordinates. One useful form is
(x−a)(x₁−a)+(y−b)(y₁−b)=r².
This is consistent with perpendicular-gradient reasoning and can reduce algebra when the contact point is known.
7. Tangents from an external point
From a point outside a circle, two tangents may be possible. Their contact points satisfy both the circle equation and a perpendicularity condition involving the centre and external point.
These problems often lead to simultaneous equations or a discriminant condition. The two possible tangents reflect the two possible contact points.
8. Tangency as a repeated intersection
A tangent line intersects the circle at exactly one point. Substituting the line equation into the circle produces a quadratic with a repeated root, so its discriminant equals zero.
This gives a second method for parameter problems involving an unknown tangent line.
9. Common mistakes
- Using the radius gradient as the tangent gradient.
- Applying the negative-reciprocal rule to a vertical radius without handling the special case.
- Finding a line through the contact point without checking perpendicularity.
- Using a tangent formula with a point that does not lie on the circle.
10. Practice
- Find the tangent to x²+y²=25 at (3,4).
- Find the tangent to (x−2)²+(y+1)²=10 at (5,0).
- Find the tangent to x²+y²=9 at (0,3).
- Explain why a tangent line substituted into a circle equation gives discriminant zero.
- Verify whether (6,2) lies on (x−2)²+(y−2)²=16 before finding its tangent.
11. Answers
- 3x+4y=25.
- Centre (2,−1); radius gradient 1/3; tangent gradient −3. y=−3x+15.
- y=3.
- There is exactly one intersection, so the resulting quadratic has a repeated root.
- Yes: 16+0=16. Radius is horizontal, so tangent is x=6.
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