Question

The polars of a point $P$ with respect to two fixed circles meet in the point $Q$. Prove that the circle on $PQ$ as diameter passes through two fixed points, and cuts both the given circles at right angles.

Answer 1

Taking the line joining the centre of the two given circles as the axis of $x$ and the radical axis as axis of $y$ (as radical axis is perpendicular to the line of centres), the equation of the circles can be written as

$x^2+y^2-2gx+c=\mathrm{0........}\left(1\right)$

$x^2+y^2-2g_{1}x+c=\mathrm{0........}\left(2\right)$

Let the point $P$ be $(h,k)$

The polars of $(h,k)$ with respect to $\left(1\right)$ and $\left(2\right)$ are

$xh+yk-g(x+h)+c=0$

or, $x(h-g)+yk+c-gh=\mathrm{0...........}\left(3\right)$

$xh+yk-g_1(x+h)+c=0$

or, $x(h-g_1)+yk+c-g_{1}h=\mathrm{0...........}\left(4\right)$

Solving $\left(3\right)$ and $\left(4\right)$, we get the co-ordinates of $Q$ as

$(-h,\frac{h^2-c}{k})$

The equation of the circle having $PQ$ as diameter will be

$(x-h)(x+h)+(y-k)(-y.\frac{h^2-c}{k})=0$

or, $k{x}^2+k{y}^2-y(k^2+h^2-c)-ck=\mathrm{0.............}\left(5\right)$

The equation is always satisfied by $x=+\sqrt{c},y=0$ and $x=-\sqrt{c},y=0$.

So the circle passes always through these points which are fixed points.

Again $\left(5\right)$ is same as $x^2+y^2-y.\frac{k^2+h^2-c}{k}-c=\mathrm{0.......}\left(6\right)$

Two circles cut orthogonally if $2g{g}^{\prime }+2f{f}^{\prime }-c-c^{\prime }=0$

Putting the values $g,g^{\prime }$ etc. from $\left(1\right)$ and $\left(6\right)$, we get

L.H.S $=2g.0+2.0(-\frac{k^2+h^2-c}{k})=c-c=0$

So the $y$ cut orthogonally. Similarly for $\left(2\right)$.