# Find the equation of normal for the circle, $x^2+y^2+2gx+2fy+c=0$ at $(x_1,y_1)$

$\begin{array}{1 1}(a)\;\large\frac{(y-y_1)}{(f+y_1)}=\frac{(x-x_1)}{(g+x_1)}\\(b)\;\large\frac{(y-y_1)}{(f-y_1)}=\frac{(x+x_1)}{(g-x_1)}\\(c)\;\large\frac{(2y-y_1)}{(f+y_1)}=\frac{(2x-x_1)}{(g+x_1)}\\(d)\;\text{None of these}\end{array}$

We know slope of tangent =$-\large\frac{(g+x_1)}{(f+y_1)}$
Hence slope of normal =$-\large\frac{f+y_1}{g+x_1}$
Hence equation of normal
$(y-y_1)=\large\frac{f+y_1}{g+x_1}$$(x-x_1) \large\frac{(y-y_1)}{(f+y_1)}=\frac{(x-x_1)}{(g+x_1)} Hence (a) is the correct answer. or Given equation of the circle is x^2+y^2+2gx+2fy+c=0 Differentiating both the sides with respect to x we get 2x+2y.\large\frac{dy}{dx}$$+2g+2f.\large\frac{dy}{dx}$$=0 \Rightarrow\:\large\frac{dy}{dx}=-\large\frac{x+g}{y+f} Slope of tangent at any point (x_1,y_1) to any curve is \large\frac{dy}{dx} at (x_1,y_1) i.e., Slope of tangent is - \large\frac{x_1+g}{y_1+f} Normal is \perp to tangent \therefore Slope of normal is -\large\frac{1}{dy/dx} =\large\frac{y_1+f}{x_1+g} \therefore Equation of normal at (x_1,y_1) is y-y_1=\large\frac{y_1+f}{x_1+g}$$(x-x_1)$
$\Rightarrow\:\large\frac{(y-y_1)}{(f+y_1)}=\frac{(x-x_1)}{(g+x_1)}$
Hence (a) is the correct answer.
edited Mar 19, 2014