The polar of the point $P \left(x_1, y_1\right)$ with respect to the circle $x^2 + y^2 = a^2$ meets the axes at $Q$ and $R$. Prove that the area of $\triangle OQR$ is $\dfrac{a^4}{2 x_1 y_1}$ where $O$ is the origin.
Given point: $\;$ $P \left(x_1, y_1\right)$
Given circle: $\;$ $x^2 + y^2 = a^2$
The polar of $P \left(x_1, y_1\right)$ with respect to the circle $\;$ $x^2 + y^2 = a^2$ $\;$ is
$x x_1 + y y_1 = a^2$ $\;\;\; \cdots \; (1)$
Let equation $(1)$ meet the $X$ axis at $Q$ and the $Y$ axis at $R$.
Then, $\;$ $Q = \left(\dfrac{a^2}{x_1}, 0\right)$ $\;$ and $\;$ $R = \left(0, \dfrac{a^2}{y_1}\right)$
Area of $\triangle OQR$
$= \dfrac{1}{2} \times OQ \times OR$
$= \dfrac{1}{2} \times \dfrac{a^2}{x_1} \times \dfrac{a^2}{y_1}$
$= \dfrac{a^4}{2 x_1 y_1}$