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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?
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  • $\begingroup$ Multiply the second equation by $k$ and add it to the first equation. Any point satisfying the two given equations (and therefore in the intersection of the two circles) must satisfy the new equation produced. To do the converse, assume that $Ax^2+Ay^2+Mx+Ny+P=0$ is a circle passing through the intersection. Subtract the second equation multiplied by $A$ from this. Then subtract the first equation multiplied by $A$ from this. Both times you get a linear equation (no quadratic terms). Since both linear pass through the same two intersection points, they must be the same. $\endgroup$ – user553213 Apr 25 '18 at 10:43
  • $\begingroup$ That technique of conveniently adding equations such that the higher degree terms cancel, which we used to get the converse, is called elimination. Maybe you have seen it in work for linear systems of equations in which you subtract equations or substitute equations in others such that some variables disappear. It is the main algorithm to study the set of solutions of polynomial systems of equations. Maybe this idea is part of what the exercise is trying to teach. $\endgroup$ – user553213 Apr 25 '18 at 10:58
  • $\begingroup$ Not likely to be used in calculus, but also the exercise gave you the equations of all the coaxal circles to the given ones. That is a notion useful in projective geometry. The book might have inherited the exercise from the times in which more attention was paid to geometry in schools. $\endgroup$ – user553213 Apr 25 '18 at 11:03
  • $\begingroup$ @deyore Thank you much. If I follow correctly, if $y=f(x)$ and $y=g(x)$ intersect, then $y+k y=f(x)+k g(x)$ or $y=\frac{f(x) + k g(x)}{k+1}$ necessarily shares the same intersection points. Presumably $A=k+1$, $M=A_1+k A_2$, $N=B_1+k B_2$ and $P=C_1+k C_2$, but this does not matter when showing the converse. $\endgroup$ – sgeos Apr 26 '18 at 7:56
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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 $k$.

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  • $\begingroup$ The other point being that you have a line instead of a circle when $k=-1$. This line does, however, go through the intersection points. $\endgroup$ – sgeos Apr 25 '18 at 10:59
  • $\begingroup$ @sgeos Yes, that observation is key. Combined with the fact that by two points passes only one line is what allows you to get the converse implication (that those are all the circles passing through the two points). $\endgroup$ – user553213 Apr 25 '18 at 11:07
  • $\begingroup$ @sgeos: sorry, I haden't completed the question. Done now. $\endgroup$ – Yves Daoust Apr 25 '18 at 11:45
  • $\begingroup$ @Yves Daoust Thank you much. Your answer is concise and makes sense. $\endgroup$ – sgeos Apr 26 '18 at 7:57
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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.

Bipolar Coords

We can learn much by googling the italicized items.

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  • $\begingroup$ Thank you much. It seemed like there was a lot to this problem beneath the surface. According to Wikipedia, bipolar coordinates can be used to solve partial differential equations. $\endgroup$ – sgeos Apr 26 '18 at 8:00

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