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 normal to the circle $x^2 + y^2 = a^2$ at any point $(x_1, y_1)$ on the circle passes through the center $(0, 0)$.Since the no

Vedclass · Accepted Answer

$x - y = 0$

Vedclass · Answer

(C) The equation of the normal to the circle $x^2 + y^2 = a^2$ at any point $(x_1, y_1)$ on the circle passes through the center $(0, 0)$.Since the normal passes through the center $(0, 0)$ and the point $P = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$,its slope $m$ is given by $m = \frac{y_1 - 0}{x_1 - 0} = \frac{1/\sqrt{2}}{1/\sqrt{2}} = 1$.The equation of the line passing through $(0, 0)$ with slope $m = 1$ is $y - 0 = 1(x - 0)$,which simplifies to $y = x$ or $x - y = 0$.

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

Similar Questions

At which point does the line $x = 0$ touch the circle $x^2 + y^2 - 2x - 6y + 9 = 0$?

$3x + 4y - 43 = 0$ is a tangent to the circle $S \equiv x^2 + y^2 - 6x + 8y + k = 0$ at a point $P$. If $C$ is the centre of the circle and $Q$ is a point which divides $CP$ in the ratio $-1:2$,then the power of the point $Q$ with respect to the circle $S = 0$ is

The equation of the common tangent touching the circle $(x-3)^2+y^2=9$ and the parabola $y^2=4x$ above the $X$-axis is

The angle between the tangents to the circle $x^2+y^2=25$ from the point $(1,7)$ is

The equation of the normal at the point $(4, -1)$ of the circle $x^2 + y^2 - 40x + 10y = 153$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module