Question

If the polar of the point $(x_1,y_1)$ with respect to the circle $x^2+y^2=a^2$ touches the circle $(x-a{)}^2+y^2=a^2$, prove that the locus of the point $(x_1,y_1)$ is $y^2+2ax-a^2=0$.

Answer 1

The polar of the point $(x_1,y_1)$ with respect to the circle $x^2+y^2-a^2=0$ is
$x{x}_1+y{y}_1-a^2=0.....\left(1\right)$
Since $\left(1\right)$ touches the circle $(x-a{)}^2+y^2=a^2$, we must have distance from the centre $(a,0)$ equal to radius $a$, i.e. $p=r$
$\frac{a{x}_1+0.y_1-a^2}{\sqrt{(x_1^2+y_1^2)}}=\pm a$
or $x_1^2+y_1^2=(x_1-a{)}^2$
or $y_1^2+2a{x}_1-a^2=0$.
The locus of $(x_1,y_1)$ is $y^2+2ax-a^2=0$.