If the pole of a straight line with respect to the circle $x^2 + y^2 = r^2$ lies on the circle $x^2 + y^2 = k^2 r^2$, prove that the line will touch the circle $x^2 + y^2 = \dfrac{r^2}{k^2}$.
Let the straight line be $\;\;$ $ax + by + c = 0$ $\;\;\; \cdots \; (1)$
Equation of circle $\;\;$ $x^2 + y^2 = r^2$ $\;\;\; \cdots \; (2)$
Let $\left(h, k\right)$ be the pole of equation $(1)$ with respect to equation $(2)$.
Then, equation $(1)$ must be identical with the polar of $\left(h, k\right)$ with respect to $(2)$.
i.e. $\;$ with $\;$ $xh + yk = r^2$
i.e. $\;$ with $\;$ $xh + yk - r^2 = 0$ $\;\;\; \cdots \; (3)$
Given: $\;$ $\left(h, k\right)$ lies on the circle $\;$ $x^2 + y^2 = k^2 r^2$
$\therefore \;$ We have, $\;$ $h^2 + k^2 = k^2 r^2$
$\implies$ $h^2 = k^2 \left(r^2 - 1\right)$ $\;\;\; \cdots \; (4)$
Now, equation $(3)$ can be written as
$y = \dfrac{-h}{k} x + \dfrac{r^2}{k}$ $\;\;\; \cdots \; (5)$
Comparing equation $(5)$ with the standard equation of line $\;$ $y = mx + c_1$ $\;$ gives
slope $= m = \dfrac{-h}{k}$ $\;\;\; \cdots \; (6a)$
and $\;$ intercept $= c_1 = \dfrac{r^2}{k}$ $\;\;\; \cdots \; (6b)$
Given equation of circle $\;\;$ $x^2 + y^2 = \dfrac{r^2}{k^2}$ $\;\;\; \cdots \; (7)$
Comparing equation $(7)$ with the standard equation of circle $\;$ $x^2 + y^2 = R^2$ $\;$ gives $\;\;$ $R^2 = \dfrac{r^2}{k^2}$
The line $\;$ $y = mx + c_1$ $\;$ will touch the circle $\;$ $x^2 + y^2 = R^2$ $\;$ if $\;\;$ $c_1^2 = R^2 \left(1 + m^2\right)$
$\therefore \;$ Equation $(5)$ will touch equation $(7)$
if $\;$ $\left(\dfrac{r^2}{k}\right)^2 = \dfrac{r^2}{k^2} \left(1 + \dfrac{h^2}{k^2}\right)$
i.e. $\;$ if $\;$ $\dfrac{r^4}{k^2} = \dfrac{r^2}{k^2} + \dfrac{r^2 h^2}{k^4}$ $\;\;\; \cdots \; (8)$
Now, in view of equation $(4)$, equation $(8)$ becomes,
if $\;$ $\dfrac{r^4}{k^2} = \dfrac{r^2}{k^2} + \dfrac{r^2}{k^4} \times k^2 \left(r^2 - 1\right)$
i.e. $\;$ if $\;$ $\dfrac{r^4}{k^2} = \dfrac{r^2}{k^2} + \dfrac{r^2}{k^2} \left(r^2 - 1\right)$
i.e. $\;$ if $\;$ $\dfrac{r^4}{k^2} = \dfrac{r^2}{k^2} + \dfrac{r^4}{k^2} - \dfrac{r^2}{k^2}$
i.e. $\;$ if $\;$ $\dfrac{r^4}{k^2} = \dfrac{r^4}{k^2}$ $\;\;$ which is true.
$\implies$ Equation $(5)$ touches circle $(7)$.
But, equations $(5)$ and $(1)$ are identical.
$\implies$ Equation $(1)$ touches circle $(7)$.
Hence proved.