Question

Find the pole of the straight line $48x+54y+53=0$ with respect to the circle $x^2+y^2-4x+3y-1=0$

Answer 1

Equation of the circle is $x^2+y^2-4x+3y-1=0$ ....(1)

and the equation of the polar is $48x+54y+54=0$ ....(2)

Let $(\alpha ,\beta )$ be the pole of line $\left(2\right)$ wrt to the circle $\left(1\right)$.

For a circle with the general equation $x^2+y^2+2gx+2fy+c=0$, the equation of the polar of pole $(\alpha ,\beta )$ is given as:

$\alpha x+\beta y+g(x+\alpha )+f(y+\beta )+c=0$ ....(3)

Therefore, polar for point $(\alpha ,\beta )$ for circle (2) is $\alpha x+\beta y-2(x+\alpha )+\frac{3}{2}(y+\beta )-1=0\text{or}(\alpha -2)x+(\beta +\frac{3}{2})y+\frac{3}{2}\beta -2\alpha -1=0$ ....(4)

Lines (2) and (4) are same, therefore

$\frac{\alpha -2}{48}=\frac{\beta +\frac{3}{2}}{54}=\frac{\frac{3}{2}\beta -2\alpha -1}{53}$ ....(5)

Hence, $27\alpha -24\beta =180$ ....(6)

$149\alpha -72\beta =58$ ....(7)

Solving equation (6) and (7), we get

$(\alpha ,\beta )=(\frac{-53}{17},\frac{-987}{136})$