Question

The equation of the normal to the circle $x^2+y^2=9$ at the point $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ is

• A. $x-y=\frac{\sqrt{2}}{3}$
• B. $x-y=0$
• C. $xy=0$
• D. $noneofthese$

Answer 1

The correct option is B $x-y=0$

$normalalwayspassesthroughcentreofcircle(0,0)\therefore e{q}^n\left(\frac{y-0}{x-0}\right)=\left(\frac{(\frac{1}{\sqrt{2}}-0)}{(\frac{1}{\sqrt{2}}-0)}\right)\left(\frac{y}{x}\right)=1ory=xorx-y=0\ast wronganswerpointshouldbe\left(\right(\frac{3}{\sqrt{2}}),(\frac{3}{\sqrt{2}}\left)\right)$ in the question.