The equation of the normal to the circle x^2+ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The equation of the circle is $x^2+y^2=16$,which has its center at $C(0,0)$.Any normal to a circle always passes through its center.Therefore,the

Vedclass · Accepted Answer

$x-y=0$

Vedclass · Answer

(C) The equation of the circle is $x^2+y^2=16$,which has its center at $C(0,0)$.Any normal to a circle always passes through its center.Therefore,the normal at the point $P\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$ is the line passing through $C(0,0)$ and $P\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.The slope $m$ of the line $CP$ is given by $m = \frac{\frac{1}{\sqrt{3}} - 0}{\frac{1}{\sqrt{3}} - 0} = 1$.The equation of the line passing through $(0,0)$ with slope $m=1$ is $y - 0 = 1(x - 0)$,which simplifies to $x - y = 0$.Thus,option $C$ is correct.

The equation of the normal to the circle $x^2+y^2=16$ at the point $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$ is

Similar Questions

The equation of a line parallel to the line $3x + 4y = 0$ and touching the circle $x^{2} + y^{2} = 9$ in the first quadrant is:

The angle between the pair of tangents drawn from $(1,3)$ to the circle $x^2+y^2-2x+4y-11=0$ is

The equation of the normal to the circle $x^2+y^2+6x+4y-3=0$ at $(1,-2)$ is

If $a > 2b > 0$,then the positive value of $m$ for which $y = mx - b\sqrt{1 + m^2}$ is a common tangent to $x^2 + y^2 = b^2$ and $(x - a)^2 + y^2 = b^2$ is:

For what possible value of $p$ is the line $x \cos \alpha + y \sin \alpha = p$ a tangent to the circle $x^2 + y^2 - 2qx \cos \alpha - 2qy \sin \alpha = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module