Question

If the line $xcos\alpha +ysin\alpha =p$ cuts the circle $x^2+y^2=a^2$ in M and N then show that the circle whose diameter is MN is
$x^2+y^2-a^2=2p$

Answer 1

The required circle by $S+\lambda P=0$ is
$\therefore (x^2+y^2-a^2)+\lambda (xcos\alpha +ysin\alpha -p)=0$
or $x^2+y^2+x\left(\lambda cos\alpha \right)+y\left(\lambda cos\alpha \right)-(a^2+\lambda p)=0$
Its centre is $(-\frac{\lambda }{2}cos\alpha ,-\frac{\lambda }{2}sin\alpha )$
Since the line MN is a diameter of the required circle and hence its centre will lie on P = 0
$\therefore -\frac{\lambda }{2}cos\alpha .cos\alpha +-\frac{\lambda }{2}sin\alpha .sin\alpha -p=0$
$\frac{\lambda }{2}.1+p=0$ $\therefore \lambda =-2p$
Hence the required circle from (1) is as given.