Question

Equation of the normal to the circle $x^2+y^2-6x+8y=0$ at $(3,1)$ is

• A. $x=3$
• B. $y=1$
• C. $x+3y=0$
• D. $3x+y=0$

Answer 1

The correct option is A $x=3$

We have,

Equation of circle is,

$x^2+y^2-6x+8y=0$

On differentiating and we get,

$2x+2y\frac{dy}{dx}-6+8\frac{dy}{dx}=0$

$\Rightarrow 2y\frac{dy}{dx}+8\frac{dy}{dx}=6-2x$

$\Rightarrow \frac{dy}{dx}=\frac{6-2x}{2y+8}$

Slope of point $(3,1)$

Then,

${\left(\frac{dy}{dx}\right)}_{(3,1)}=\frac{6-2\times 3}{2\times 1+8}=0$

Equation of normal is

$y-y_1={\left(\frac{dy}{dx}\right)}_{(3,1)}(x-x_1)$

$\Rightarrow y-1={\left(\frac{dx}{dy}\right)}_{(3,1)}(x-3)$

$\Rightarrow y-1=\frac{1}{0}(x-3)$

$\Rightarrow x-3=0$

$\Rightarrow x=3$

Hence, this is the answer.