Circle parameter problems ask which values make a geometric condition true. The unknown may control a centre, radius, line, contact point or intersection pattern. The key move is to translate the geometry into an equation before solving the parameter.
1. Build the condition dictionary
- Point lies on circle → substitute coordinates into the circle equation.
- Line tangent to circle → one repeated intersection or perpendicular radius at contact.
- Two intersections → discriminant > 0 after substitution.
- No real intersection → discriminant < 0.
- Equal radii or distances → compare squared distances.
2. Parameter from a point-on-circle condition
The circle (x−k)²+y²=25 passes through (3,4). Substitute the point:
(3−k)²+16=25, so (3−k)²=9.
Thus 3−k=±3, giving k=0 or 6.
There are two possible centres on the horizontal axis, each five units from the point.
3. Parameter from a radius condition
For x²+y²−4x+6y+k=0, complete the square:
(x−2)²+(y+3)²=13−k.
If the radius is 5, then 13−k=25, so k=−12.
4. Parameter for a real circle
Using the same family, radius squared is 13−k. For a nondegenerate real circle, require 13−k>0, hence k<13.
At k=13 the radius is zero, giving only the single centre point rather than an ordinary circle of positive radius.
5. Parameter from horizontal tangency
Find k if y=k is tangent to (x−2)²+(y+1)²=16.
The centre is (2,−1) and radius is 4. Horizontal tangents lie 4 units above or below the centre, so k=3 or −5.
Substitution confirms this: (x−2)²+(k+1)²=16 must have one real x-value, requiring (k+1)²=16.
6. Parameter from line-circle tangency
Find c such that y=x+c is tangent to x²+y²=8.
Substitute y=x+c:
x²+(x+c)²=8, so 2x²+2cx+c²−8=0.
Tangency requires discriminant zero:
(2c)²−4(2)(c²−8)=0.
4c²−8c²+64=0, so c²=16 and c=±4.
7. Parameter for two intersections
For the same line family y=x+c and circle x²+y²=8, two distinct intersections require the discriminant to be positive:
64−4c²>0, so c²<16 and −4<c<4.
8. Geometric distance can replace discriminant algebra
A line Ax+By+C=0 is tangent to a circle when the perpendicular distance from the centre to the line equals the radius.
For a circle centred at the origin with radius r, tangency of Ax+By+C=0 requires |C|/√(A²+B²)=r.
This gives an independent geometric route and a useful check on discriminant work.
9. Parameter from equal circle distances
Suppose P(k,0) is equidistant from centres A(−2,3) and B(4,3). Compare squared distances:
(k+2)²+9=(k−4)²+9.
Thus (k+2)²=(k−4)², giving k=1. This is the perpendicular-bisector condition in coordinate form.
10. Common mistakes
- Solving for an intersection coordinate when the question asks for a parameter range.
- Using discriminant zero before substituting the line into the circle.
- Forgetting that a square equation can produce two parameter values.
- Treating radius squared as radius.
- Ignoring the condition that a real circle must have positive radius squared.
11. Practice
- Find k if (x−k)²+y²=9 passes through (0,3).
- For x²+y²+2x−4y+k=0, find k if the radius is 4.
- Find k if y=k is tangent to x²+y²=25.
- Find c if y=2x+c is tangent to x²+y²=5.
- For what c does y=x+c cut x²+y²=2 at two distinct points?
12. Answers
- k=0, because k²+9=9.
- Centre (−1,2); radius squared 5−k=16, so k=−11.
- k=±5.
- Substitute y=2x+c: 5x²+4cx+c²−5=0. Tangency gives 16c²−20(c²−5)=0, so c²=25 and c=±5.
- 2x²+2cx+c²−2=0. Require 4c²−8(c²−2)>0, so 16−4c²>0 and −2<c<2.
Continue with Intersections of Lines and Circles, Finding Tangents to Circles, or return to the Additional Mathematics Hub.