Question

Find the pole of the straight line 2x + y + 12 = 0 with respect to the circle $x^2+y^2-4x+3y-1=0.$

Answer 1

Polar of the point $(x_1,y_1)$ with respect to the circle $x^2+y^2-4x+3y-1=0$ is given by $x{x}_1+y{y}_1-2x-2x_1+1.5y+1.5y_1-1=0$
Comparing the above equation with the equation of the straight line $2x+y+12=0$, we have
$\frac{x_1-2}{2}=\frac{y_1+1.5}{1}=\frac{-2x_1+1.5y_1-1}{12}$
$\Rightarrow 6x_1-12+2x_1-1.5y_1+1=0$
or $8x_1-1.5y_1-11=0...\left(1\right)$
Also, $12y_1+18+2x_1-1.5y_1+1=0$ or $2x_1+10.5y_1+19=0...\left(2\right)$
Multiplying equation $\left(1\right)$ by $7$ and adding that to equation $\left(2\right)$, we have
$58x_1=58$ or $x_1=1$
$\therefore y_1=-2$
The pole is thus $(1,-2)$