To prove that the centre of 2 circles and the two points at which the 2 circles cut and the origin lie on a circle.

Question

Let the circles
$$x^2+y^2-2cy-a^2=0~~~~and~~~~x^2+y^2-2bx+a^2=0$$
with centres at $A$ and $B$ intersect at $P$ and $Q$. Show that the points $A,B,P,Q$ and $O=(0,0)$ lie on a circle.

My work:
I found out the centre of the 2 circles to be $A(0,c),B(b,0)$ and also found out that the points at the which circle cut, $P,Q$ are the common chord of both the circles. So,I found out the equation of the chord of the circle to be $bx-cy=a^2$. Now, I am stuck. I tried to figure out that these points are concylic by considering two quadrilaterals formed by these points but nothing to progress. Please help!

Answer 1

Let $\gamma$ be the circle with diameter $\overline{AB}$. Obviously, $\gamma$ contains both $A$ and $B$. Because $\angle AOB$ is a right angle, point $O$ is also on $\gamma$ by a converse of the Inscribed Angle Theorem (more specifically, of Thales' Theorem).

Writing the circle equations in a more useful form $$\begin{align} x^2 + ( y - c )^2 &= a^2 + c^2 \\ ( x - b )^2 + y^2 &= b^2 - a^2 \end{align}$$ we see that $|\overline{OA}| = |c|$ and $|\overline{OB}| = |b|$, and that $$|\overline{AP}|^2 = |\overline{AQ}|^2 = a^2 + c^2 \qquad\qquad |\overline{BP}|^2 = |\overline{BQ}|^2 = b^2 - a^2$$

From right triangle $\triangle AOB$: $$|\overline{AB}|^2 = |\overline{OA}|^2 + |\overline{OB}|^2 = b^2 + c^2$$

By interesting coincidence, $$|\overline{AP}|^2 + |\overline{BP}|^2 = (a^2+c^2)+(b^2-a^2) = b^2 + c^2 = |\overline{AB}|^2$$ so that $\triangle ABP$ ---likewise, $\triangle ABQ$--- is also a right triangle. Consequently, $P$ and $Q$ are on $\gamma$ by the converse of the Thales' Theorem.

Answer 2

Hint: By what you have found the center must be at $K=(b/2,c/2)$ since this is where the perpendicular bisectors of $OA$ and $OB$ meet. So far a circle with this center goes through the three points $O,A,B.$ Now you only need solve for the two circle intersections, and check their distance to $K$ is the same as from $O$ to $K$.

Added: It turns out the intersections of the two circles are a bit of a mess to find. However it turns out we don't need their expressions in order to prove all five points are concyclic.

Let the circle equations be labeled by their centers: $$A :\ x^2+y^2-2cy-a^2=0, \\ B: \ x^2+y^2 -2bx+a^2=0.\tag{1}$$ Now $B=0$ can be rewritten as $(x-b)^2+y^2=b^2-a^2,$ so that one must assume the right side of this is positive in order that $B=0$ denote an actual circle. We'll assume that. For the intersection of the two circles $A,B$ we would solve $A=0,B=0$ simultaneously. This is the same as solving $(A+B)/2=0,\ (A-B)/2=0$ simultaneously. We have $$(A+B)/2=0:\ \ x^2+y^2-bx-cy=0, \\ (A-B)/2=0: \ \ bx-cy-a^2=0. \tag{2}$$

This pair of equations (when set to $0$) is really not easier than the direct pair $A,B$. But an interesting thing happens. If we take the circle through $A(0,c),B(b,0),O(0,0)$ we use its center $(b/2,c/2)$ and squared radius $b^2/4+c^2/4$, and find the equation of the circle through $A,B,O$ to be none other than $x^2+y^2-bx-cy=0$, which is one of the two equations of system $(2)$ we were to solve to determine where the two circles given by equations $A=0,B=0$ intersect each other. This proves that, even without getting the specific coordinates of these two other points $P,Q$, they must lie on the circle through $A,B,O.$ Conclusion: all five points on the same circle.

Answer 3

I took a slightly different tack after seeing a graph of the situation. If we simply subtract one circle equation from the other, we obtain

$$ 2bx \ - \ 2cy \ - \ 2a^2 \ = \ 0 \ \ , $$

which is the equation of the line through $ \ P \ $ and $ \ Q \ , $ since this is telling us something about the intersections of the two circles. (And, I agree, we don't really need to know the coordinates of those points.) This line, containing the chord $ \ PQ \ $ in both circles, has a slope of $ \ \frac{b}{c} \ . $ But the line through the circle centers $ \ A \ $ and $ \ B \ $ has slope $ \ - \frac{c}{b} \ , $ so these two lines are perpendicular. This means that the line $ \ AB \ $ contains a diameter of each circle, and potentially the one we seek.

Now since $ \ A \ $ and $ \ B \ $ lie on this putative common circle, one might simply choose the segment $ \ AB \ $ to be a diameter, which would put the center of this circle at the midpoint $ \ ( \frac{b}{2} \ , \ \frac{c}{2} ) $ . If this circle is to pass through the origin, the Pythagorean Theorem tells us that $ \ r^2 \ = \ \frac{b^2}{4} + \frac{c^2}{4} \ . $

We have thereby described a circle through $ \ O \ , \ A \ , \ $ and $ \ B \ , \ (x - \frac{b}{2} )^2 \ + \ (y - \frac{c}{2} )^2 \ = \ \frac{b^2}{4} + \frac{c^2}{4} \ . $ Does it pass through $ \ P \ $ and $ \ Q \ $ ? This circle equation can be re-written as $ \ x^2 + y^2 \ = \ bx + cy \ . $ Inserting this into the two circle equations to replace their first two terms yields

$$ bx \ + \ cy \ - \ 2cy \ - \ a^2 \ = \ 0 \ \ \text{and} \ \ bx \ + \ cy \ - \ 2bx \ + \ a^2 \ = \ 0 $$

$$ \Rightarrow \ \ bx \ - \ cy \ - \ a^2 \ = \ 0 \ \ \text{and} \ \ -bx \ + \ cy \ + \ a^2 \ = \ 0 . $$

These are both exactly the equation for the line containing the chord $ \ PQ \ $ (after a scalar multiplication). So this chord passes through the intersections of all three circles and thus the new circle passes through all five points.

[I am proposing this solution not because it is fundamentally distinct from coffeemath's, but only because it uses a couple different geometrical facts. Either way, the analytical portion is simple enough that I imagine there is probably a "classical" geometric proof.]