<s>[INST] 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. 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!
 [/INST] Let $C_1$ be the circle defined by the equation $x^2+y^2-2cy-a^2=0$ and $C_2$ be the circle defined by the equation $x^2+y^2-2bx+a^2=0.$ Now $$x^2+y^2-2cy-a^2=0 \\ \implies (x-0)^2+(y-c)^2=\left(\sqrt{a^2+c^2}\right)^2 \text{and}\\ x^2+y^2-2bx+a^2=0\\ \implies (x-b)^2+(y-0)^2=\left(\sqrt{b^2-a^2}\right)^2.$$ This implies that $C_1$ is a circle with it's centre as $A=(0,c)$ and radius $R_1=\sqrt{a^2+c^2}$ and $C_2$ is a circle with it's centre as $A=(b,0)$ and radius $R_2=\sqrt{b^2-a^2}.$ 
Now in $\Delta AQB$, we have $$AB^2=(b-0)^2+(0-c)^2=b^2+c^2, \\ AQ^2=R_1^2=a^2+c^2 \text{ and } BQ^2=R_2^2=b^2-a^2, \\ \implies AB^2=AQ^2+BQ^2.$$ 
Therefore $\Delta AQB$ is a right-angled triangle with $\angle AQB=90^0.$ 
Now we know that $\Delta AQB \cong \Delta APB,$ which implies that $\angle APB=90^0$. 
Therefore, in quadrilateral $AQBP$, we have $\angle APB+\angle AQB=180^0,$ which implies that $AQBP$ is a cyclic quadrilateral. This implies that points $A,Q,B,P$ lie on a circle having $AB$ as the diameter. 
Now, we also have $AOB=90^0,$ which implies that $O$ lies on the same circle defined by the points $A,Q,B,P$. 
Hence, the points $A,B,P,Q$ and $O$ lie on a circle.  
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