Proof for points of intersecting circles.

Question

I am an adult software developer who is trying to do a math reboot. I am working through the exercises in the following book.

Ayres, Frank , Jr. and Elliott Mendelson. 2013. Schaum's Outlines Calculus Sixth Edition (1,105 fully solved problems, 30 problem-solving videos online). New York: McGraw Hill. ISBN 978-0-07-179553-1.

Most of the precalculus problems are easy, but every chapter seems to have a problem that is significantly harder than the others. The following problem struck me as especially hard. Frankly, it seems completely out of depth compared to the other problems. Clearly I have a gap in my knowledge.

Chapter 4 circles, problem 24.

Let $\mathscr{C}_1$ and $\mathscr{C}_2$ be two intersecting circles determined by the equations $x^2+y^2+A_1x+B_1y+C_1=0$ and $x^2+y^2+A_2x+B_2y+C_2=0$. For any number $k \ne -1$, show that $$ x^2+y^2+A_1x+B_1y+C_1+k(x^2+y^2+A_2x+B_2y+C_2)=0 $$ is the equation of a circle through the intersection of $\mathscr{C}_1$ and $\mathscr{C}_2$. Show, conversely, that every such circle may be represented by such an equation for a suitable $k$.

My attempt.

I spent weeks with SageMath and WolframAlpha figuring out that the intersecting points $P$ and $Q$ of $\mathscr{C}_1$ and $\mathscr{C}_2$ given $A_1$, $B_1$, $C_1$, $A_2$, $B_2$ and $C_2$ are as follows.

$$ P_x = \frac{ (A_1 - A_2)(C_2 - C_1) - (B_1 - B_2) \bigg(\frac{A_2 B_1 - A_1 B_2}{2} - \sqrt{ \frac{(A_2 B_1 - A_1 B_2)^2}{4} + (A_1 - A_2)(A_2 C_1 - A_1 C_2) + (B_1 - B_2)(B_2 C_1 - B_1 C_2) - (C_1 - C_2)^2 }\bigg)}{ ( A_1 - A_2 )^2 + ( B_1 - B_2 )^2 } \\ P_y = \frac{ (B_1 - B_2)(C_2 - C_1) - (A_1 - A_2)\bigg(\frac{A_1 B_2 - A_2 B_1}{2} + \sqrt{ \frac{(A_2 B_1 - A_1 B_2)^2}{4} + (A_1 - A_2) (A_2 C_1 - A_1 C_2) + (B_1 - B_2) (B_2 C_1 - B_1 C_2) - (C_1 - C_2)^2 }\bigg)}{ ( A_1 - A_2 )^2 + ( B_1 - B_2 )^2 } \\ Q_x = \frac{ (A_1 - A_2)(C_2 - C_1) - (B_1 - B_2)\bigg(\frac{A_2 B_1 - A_1 B_2}{2} + \sqrt{ \frac{(A_2 B_1 - A_1 B_2)^2}{4} + (A_1 - A_2)(A_2 C_1 - A_1 C_2) + (B_1 - B_2)(B_2 C_1 - B_1 C_2) - (C_1 - C_2)^2 }\bigg)}{ ( A_1 - A_2 )^2 + ( B_1 - B_2 )^2 } \\ Q_y = \frac{ (B_1 - B_2) (C_2 - C_1) - (A_1 - A_2) \bigg(\frac{A_1 B_2 - A_2 B_1}{2} - \sqrt{ \frac{(A_2 B_1 - A_1 B_2)^2}{4} + (A_1 - A_2) (A_2 C_1 - A_1 C_2) + (B_1 - B_2) (B_2 C_1 - B_1 C_2) - (C_1 - C_2)^2 }\bigg)}{ ( A_1 - A_2 )^2 + ( B_1 - B_2 )^2 } $$

I then found that the intersecting points of $\mathscr{C}_1$ or $\mathscr{C}_2$ and $x^2+y^2+A_1x+B_1y+C_1+k(x^2+y^2+A_2x+B_2y+C_2)=0$ are the same as the above. $k$ was "magically" simplified out. I am convinced that the assertion in the question is correct, but I do not feel that I properly solved the problem.

I have three questions.

1. What does a proper proof that solves the problem look like?
2. What are effective ways to approach problems like this?
3. What is this precalculus warm up problem trying to teach me as I move towards calculus?

Answer 1

A circle is a conic such that the coefficients of $x^2$ and $y^2$ are equal and there is no $xy$ term.

• As the combined equation has this property, it does describe a circle.

• As $0+k\,0=0$, any point that satisfies the equation of both circles satisfies the combined equation, hence all these circles share the same intersections (if any).

---

Conversely,

plugging the coordinates of the two intersection points shows that there are two (affine) relations between the coefficients and you can eliminate two of them, let $A$ and $B$. Then any given $C$ can be reconstructed as a combination $C_1+kC_2$, and compatible $A$ and $B$ will follow. Hence any circle in the pencil can be obtained by a linear combination of two distinct others.

---

Geometric addendum:

By symmetry, circles that share two points all have their center on the mediatrix of these points. If you choose the center, the radius follows and the family of circles has a single degree of freedom, which can be conveyed by the single parameter $k$.

Answer 2

I think it is beneficial to grasp the beautiful topic of Apollonian Circles and the bigger picture of Radical axes/Co-axal circles in analytical geometry than time out with sage or WA at the present stage of inquiry.

By subtraction of circle equations you get a radical axis of two equal tangents for each of two intersecting circles ($k=-1$ linear equation, straight line).

By their addition with a linear constant $k+1 \ne 0 $ you get a radical axis from which you can draw set of four equal tangents, with all circles having the same $x,y$ coefficients.

The two sets are set together in Bipolar coordinates. In the intersectiong set angle contained in a circle is constant parameter. In the disjunct set ratio of segments is a constant parameter for the moving apex point on circle as constant ratio of connecting sides.

We can learn much by googling the italicized items.