The slope of the normal to the circle $x^2+y^2+2 g x+2 f y$ $+\mathrm{c}=0$ at $\left(\mathrm{x}_1,…
The slope of the normal to the circle $x^2+y^2+2 g x+2 f y$ $+\mathrm{c}=0$ at $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is
$-\left(\frac{x_1+g}{y_1+f}\right)$
$-\left(\frac{y_1+f}{x_1+g}\right)$
$\frac{x_1+g}{y_1+f}$
$\frac{y_1+f}{x_1+g}$
Solution
Given equation of circle is
$
x^2+y^2+2 g x+2 f y+c=0
$
$\therefore$ centre $\mathrm{c}(-\mathrm{g},-\mathrm{f})$
Slope of normal $C P=\frac{y_1+f}{y_1+g}$