Question

Find the equation of the normal at $P(3,5)$ of the circle $S\equiv x^2+y^2-10x-2y+6$

Answer 1

Given equation of circle-

$S:x^2+y^2-10x-2y+6=0$

here $g=-5,f=-1$

As we know that, the equation of normal to the circle at $(x_1,y_1)$ is-

$\frac{x-x_1}{x_1+g}=\frac{y-y_1}{y_1+f}$

Therefore,

Equation of normal at $(3,5)$ to the circle $S$ is-

$\frac{x-3}{3+(-5)}=\frac{y-5}{5+(-1)}$

$\Rightarrow 4(x-3)=-2(y-5)$

$\Rightarrow 4x-12=-2y+10$

$\Rightarrow 2x+y-11=0$

Hence the equation of normal to the given circle at $(3,5)$ is $2x+y-11=0$.