The pole of the straight line $9 x+y-28=0$ with respect to the circle $2 x^2+2 y^2-3 x+5 y-7=0$ is

Question

Aicharya · Accepted Answer

Let $(h, k)$ be the pole $9 x+y-28=0$ with respect to given circle $2 x^2+2 y^2-3 x+5 y-7=0$
Then equation of polar is $T=0$
$\begin{aligned} & 2 h x+2 k y-\frac{3(h+x)}{2}+\frac{5(k+y)}{2}-7=0 \\ & \Rightarrow\left(2 h-\frac{3}{2}\right) x+\left(2 k+\frac{5}{2}\right) y+\frac{5 k-3 h-14}{2}=0 \end{aligned}$
On comparing with $9 x+y-28=0$, we get
$\frac{2 h-\frac{3}{2}}{9}=\frac{2 k+\frac{5}{2}}{1}=\frac{5 k-3 h-14}{-2 \times 28}$
On solving we get $(h, k)=(3,-1)$.

The pole of the straight line $9 x+y-28=0$ with respect to the circle $2 x^2+2 y^2-3 x+5 y-7=0$ is

Solution