Let P be any point on the circle $x^2+y^2=25$. Let L be the chord of contact of P with respect to the circle…
Let P be any point on the circle $x^2+y^2=25$. Let L be the chord of contact of P with respect to the circle $x^2+y^2=$ 9. The locus of the poles of the lines L with respect to the circle $x^2+y^2=36$ is
$y^2=20 x$
$\frac{x^2}{9}+\frac{y^2}{36}=1$
$x^2+y^2=400$
$\frac{x^2}{25}-\frac{y^2}{16}=1$
Solution
Let $P(r, s)$ be point on circle $x^2+y^2=25$
$\mathrm{r}^2+\mathrm{s}^2=25...(i)$
Equation of chord of contact of $P$ w.r.t. circle $x^2+y^2=9$ is L
$\mathrm{L}: \mathrm{xr}+\mathrm{ys}-9=0...(ii)$
Poles of line L w.r.t. circle $x^2+y^2=36$ is ( $h, k$ )
then $\mathrm{xh}+\mathrm{yk}-36=0...(iii)$
Solving (ii) and (iii), we get substitute value of r and s in $e^{\mathrm{n}}$ (i)
$\frac{h}{4}=r, \frac{k}{4}=s \Rightarrow \frac{h^2}{16}+\frac{k^2}{16}=25$
So required locus of pole is $x^2+y^2=400$.