Chapter 4 · Geometry and Equations · Concept 3
4-26 Circle Through the Points of Intersection of Two Circles
If a point sits on both circles, it makes both equations zero — so it makes any mixture of them zero too. That one observation gives you an endless family of curves through the same two points, and picking the right member is a matter of choosing a single number.
Guiding question
Can you draw a new circle through two crossings without ever finding them?
Learning outcome. Find the equation of a circle passing through the points of intersection of two other circles.
Mix the two equations, then steer with k
A curve through the crossings of two circles can always be written as
k(x² + y² + px + qy + r) + (x² + y² + lx + my + n) = 0
Multiply one circle by k, add the other. Then use whatever extra fact the question gives you to find k.
Finish by making the x² and y² coefficients 1.
After k goes in, both squares carry a factor of k + 1. An answer that still reads ½x² + ½y² + x + 3y − 15/2 = 0 is not finished — multiply through by 2 to get x² + y² + 2x + 6y − 15 = 0. The book states this as a separate rule because it is separately forgotten.
Why every member works.
At a crossing point the first bracket is 0 and the second bracket is 0. So the whole expression is k × 0 + 0 = 0, whatever k happens to be. The two crossing points can never fall off the family — which is why you never have to find them.
And lesson 4-25 was hiding in here.
Put k = −1. Both x² terms cancel and both y² terms cancel, and what is left is a straight line — exactly the line you get by subtracting. The two lessons are one idea: 4-25 is this family at its flattest.
Slide k anywhere. The circle swells, shrinks and flips sides — and the two red crossing points never move. Stop at k = −1 to watch it flatten into the line from 4-25, and at k = −1/2 for the one that also catches the green point.
C₁: x² + y² − 2x − 6y + 5 = 0
C₂: x² + y² − 5 = 0
k · C₁ + C₂ = 0
k = -0.50
x² + y² + 2x + 6y − 15 = 0
Find the circle through the crossings of x² + y² − 2x − 6y + 5 = 0 and x² + y² = 5, and through the point (3, 0).
- 1
Write the family down, with k in front of one circle.
k(x2 + y2 − 2x − 6y + 5) + (x2 + y2 − 5) = 0 …(i)Every curve in this family already passes through both crossing points, whatever k turns out to be. That part is finished before we start.
Check with the extra point.
(3, 0): 9 + 0 + 6 + 0 − 15 = 0 ✓. And you never needed the crossing points themselves — the family guaranteed them from the first line. That is what makes this method worth having.
Work with a partner. Solve it separately, then compare your k before comparing your final equations. If your k values differ, one of you has an arithmetic slip in the substitution — and it is far easier to find there than in the collected answer.
Find the equation of the circle passing through the points of intersection of the two circles x² + y² − 2x − 6y + 6 = 0 and x² + y² + 4x − 2y − 11 = 0, and through the point (2, 2).
Leaving the squares with a coefficient.
After k goes in, x² and y² carry a factor of k + 1. Multiply through so both are 1 — the book asks for this in its own bullet point.
Arithmetic slips inside the brackets.
One wrong sign when substituting the extra point gives a wrong k, and every line after it is wrong too. Work each bracket out on its own.
Expecting k to be negative.
Most of the examples give a negative k, but the Challenge gives k = 1/2. There is no rule — k is whatever the extra point makes it.
Finding the crossing points first.
You never need them. The family passes through them automatically, and hunting for them is a page of work that earns nothing.
Forgetting that k = −1 is a line, not a circle.
At k = −1 the squares cancel completely. If your extra point forces k = −1, the answer is a straight line — which is lesson 4-25.
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