If the slope of the tangent of the circle S e | vedclass

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

Question

(B) Given equation of the circle is $S \equiv x^2+y^2-13=0$.On differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$,which implies $\

Vedclass · Accepted Answer

an internal point with respect to the circle $S=0$

Vedclass · Answer

(B) Given equation of the circle is $S \equiv x^2+y^2-13=0$.On differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{y}$.The slope of the tangent $m$ at $(2,3)$ is $m = \left. \frac{dy}{dx} \right|_{(2,3)} = -\frac{2}{3}$.Now,the point $\left(m, -\frac{1}{m}\right)$ becomes $\left(-\frac{2}{3}, \frac{3}{2}\right)$.To check the position of this point with respect to the circle,we substitute it into the expression $S(x, y) = x^2+y^2-13$:$S\left(-\frac{2}{3}, \frac{3}{2}\right) = \left(-\frac{2}{3}\right)^2 + \left(\frac{3}{2}\right)^2 - 13 = \frac{4}{9} + \frac{9}{4} - 13 = \frac{16 + 81 - 468}{36} = \frac{97 - 468}{36} = -\frac{371}{36}$.Since $S\left(-\frac{2}{3}, \frac{3}{2}\right) < 0$,the point lies inside the circle.

If the slope of the tangent of the circle $S \equiv x^2+y^2-13=0$ at $(2,3)$ is $m$,then the point $\left(m, \frac{-1}{m}\right)$ is

Similar Questions

The slope of the tangent to the circle $(x-6)^2 + y^2 = 2$,which passes through the focus of the parabola $y^2 = 16x$,is:

The equations of the tangent to the circle $5x^2 + 5y^2 = 1$ parallel to the line $3x + 4y = 1$ are

The equations of the tangents to the circle $x^2+y^2=36$ which are perpendicular to the line $5x+y-2=0$ are

Find the equations of the tangents to the circle $x^2 + y^2 - 6x + 4y = 12$ which are parallel to the line $4x + 3y + 5 = 0$.

$3x + 4y - 43 = 0$ 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

Vedclass Test Series

Exam Paper Generator

Online Exam Module