Question

Find equation of normal to the circle $x^2+y^2+x+y=0$ This normal passes through $(2,1)$.

Answer 1

$x^2+y^2+x+y=0$ Differntiating wrt x,

$2x+2y\left(\frac{dy}{dx}\right)+1+\frac{dy}{dx}=0$

$\left(\frac{dy}{dx}\right)(2y+1)=-1-2x$

$\therefore \frac{dy}{dx}=\frac{-(1+2x)}{1+2y}$

At (2,1) $\left(\frac{dy}{dx}\right)$= slope = $\frac{-5}{3}$

$\therefore$ Slope of normal = $\frac{3}{5}=m^{\prime }$

$\therefore$ Equation of normal $(y-y_1)=m^{\prime }(x-x_1)$

$(y-1)=\frac{3}{5}(x-2)$

3x-5y-1=0