The equation of polar of $(1,1)$ with respect to the circle $x^2+y^2+4 x+6 y-3=0$ is
The equation of polar of $(1,1)$ with respect to the circle $x^2+y^2+4 x+6 y-3=0$ is
2 x + 3y − 1 = 0
3x + 4y + 8 = 0
4x + 3y + 2 = 0
3x + 4y + 2 = 0
Solution
Given, circle is $x^2+y^2+4 x+6 y-3=0 \ldots$
Given, point is P(1,1).
Since, equation of polar form of point $\left(x_1, y_1\right)$ with respect to circle $x^2+y^2+2 g x+2 f y+c=0$ is
$x \cdot x_1+y \cdot y_1+2 g\left(\frac{x+x_1}{2}\right)+2 f\left(\frac{y+y_1}{2}\right)+c=0$
x ×1 + y ×1 + 2(x + 1) + 3(y + 1) − 3 = 0
x + y + 2x + 2 + 3y + 3 − 3 = 0
$\Rightarrow \quad 3 x+4 y+2=0$