Question

The equation of the normal to the circle $x^2+y^2=a^2$ at point $(x^{\prime },y^{\prime })$ will be:

• A. $x^{\prime }y-x{y}^{\prime }=0$
• B. $x{x}^{\prime }-y{y}^{\prime }=0$
• C. $x^{\prime }y+xy=0$
• D. $x{x}^{\prime }+y{y}^{\prime }=0$

Answer 1

The correct option is A $x^{\prime }y-x{y}^{\prime }=0$

Given, $x^2+y^2=a^2$
On differentiating w.r.t. $x$, we get
$2x+2y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(x^{\prime },y^{\prime })}=-\frac{x^{\prime }}{y^{\prime }}$
Slope of tangent is $-\frac{x^{\prime }}{y^{\prime }}$
So, Slope of normal will be $-\frac{dx}{dy}$ i.e. $\frac{y^{\prime }}{x^{\prime }}$
$\therefore$ Equation of normal is
$y-y^{\prime }=\frac{y^{\prime }}{x^{\prime }}(x-x^{\prime })$
$\Rightarrow x^{\prime }y-y^{\prime }{x}^{\prime }=x{y}^{\prime }-y^{\prime }{x}^{\prime }$
$\Rightarrow x^{\prime }y-x{y}^{\prime }=0$