Suppose the angle between the tangents drawn | vedclass

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

Question

(A) The equation of the circle is $(x+\lambda)^2+(y+1)^2=\lambda^2$.Here,the center $C$ is $(-\lambda, -1)$ and the radius $r$ is $|\lambda|$.Let $O$

Vedclass · Accepted Answer

$\lambda^2=1$

Vedclass · Answer

(A) The equation of the circle is $(x+\lambda)^2+(y+1)^2=\lambda^2$.Here,the center $C$ is $(-\lambda, -1)$ and the radius $r$ is $|\lambda|$.Let $O$ be the origin $(0,0)$ and $P$ be the point of contact of a tangent from $O$ to the circle.The angle between the tangents is $\frac{\pi}{2}$,so the angle $\angle COP = \frac{1}{2} 	imes \frac{\pi}{2} = \frac{\pi}{4} = 45^{\circ}$.In the right-angled triangle $	riangle OCP$,we have $	an(\angle COP) = \frac{CP}{OP} = \frac{r}{OP}$.Since $\angle COP = 45^{\circ}$,$	an(45^{\circ}) = 1$,so $OP = CP = |\lambda|$.Also,the distance $OC = \sqrt{(-\lambda-0)^2 + (-1-0)^2} = \sqrt{\lambda^2+1}$.By the Pythagorean theorem in $	riangle OCP$,$OC^2 = OP^2 + CP^2$.Substituting the values,$(\sqrt{\lambda^2+1})^2 = |\lambda|^2 + |\lambda|^2$.$\lambda^2 + 1 = 2\lambda^2$.Therefore,$\lambda^2 = 1$.

Suppose the angle between the tangents drawn from $(0,0)$ to the circle $(x+\lambda)^2+(y+1)^2=\lambda^2$ is $\frac{\pi}{2}$. Then,$\lambda$ satisfies

Similar Questions

The equation of the tangent at the point $\left( \frac{ab^2}{a^2 + b^2}, \frac{a^2b}{a^2 + b^2} \right)$ to the circle $x^2 + y^2 = \frac{a^2b^2}{a^2 + b^2}$ is

The equation of a circle which has a tangent $3x + 4y = 6$ and two normals given by $(x - 1)(y - 2) = 0$ is

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

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 line $lx + my + n = 0$ will be a tangent to the circle $x^2 + y^2 = a^2$ if

Vedclass Test Series

Exam Paper Generator

Online Exam Module