Question

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

• A. $y=2x$
• B. $x+y=0$
• C. $x-y=0$
• D. None of these

Answer 1

The correct option is A $y=2x$

Given equation of circle is $x^2+y^2=9$

Centre of the given circle is point $(0,0)$. Equation of normal to the circle from the point $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ can be found by two-point from the point of line.

Let $P(x_1,y_1)=(0,0)$

$Q(x_2,y_2)=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

Equation of $PQ$ is,

$y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)(x-x_1)$

$y-0=\left(\frac{\frac{1}{\sqrt{2}}-0}{\frac{1}{\sqrt{2}}-0}\right)(x-0)$

$y=x⟶$ Required equation of normal through $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ or $x-y=0$