Question

$OAB$ is any chord of a circle which passes through $O$ a point in the plane of the circle and meets it in points $A$ and $B$. A point $P$ is taken on this chord such that $OP$ is (a) A.M. (b) G.M. of $OA$ and $OB$. Prove that the locus of $P$ in either case is a circle and find the equation also.

Answer 1

Let $O$ be $(\alpha ,\beta )$ then any line through $O$ is $\frac{x-\alpha }{\cos \theta }=\frac{y-\beta }{\sin \theta }=\frac{r}{OP}(\frac{=r_1,r_2}{OAOB})$
$\therefore x=r\cos \theta +\alpha ,y=r\sin \theta +\beta .....\left(1\right)$
It meets the given circle $x^2+y^2+a^2$ in $A$ and $B$
$\therefore (r\cos \theta +\alpha {)}^2+(r\sin \theta +\beta {)}^2=a^2$
or $r^2+2r(\alpha \cos \theta +\beta \sin \theta )+({\alpha }^2+{\beta }^2-a^2)=0$
Above is a quadratic in $r$ and its roots are $OA=r_1,OB=r_2$.
(i) Given $OP$ is A.M. of $OA$ and $OB$
$\therefore 2r=r_1+r_2$
$2r=2(\alpha \cos \theta +\beta \sin \theta )$, by $\left(2\right)$
Cancel $2$ and multiply by $r$
$\therefore r^2+(\alpha r\cos \theta +\beta r\sin \theta )=0$
or $(x-\alpha {)}^2+(y-\beta {)}^2+\alpha (x-\alpha )+\beta (y-\beta )=0$, by $\left(1\right)$
It represents a circle.
(ii) Given $OP$ is G.M. of $OA$ and $OB$
$\therefore r^2=r_1{r}_2={\alpha }^2+{\beta }^2-a^2$ by $\left(2\right)$
or $(x-\alpha {)}^2+(y-\beta {)}^2={\alpha }^2+{\beta }^2-a^2$ by $\left(1\right)$
Above is also a circle.