From any point on the circle x^2 + y^2 = a^2, | vedclass

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

Question

(C) Let $P$ be any point on the circle $x^2 + y^2 = a^2$. Let $PQ$ and $PR$ be the tangents drawn from $P$ to the circle $x^2 + y^2 = a^2 \sin^2 \alph

Vedclass · Accepted Answer

$2\alpha$

Vedclass · Answer

(C) Let $P$ be any point on the circle $x^2 + y^2 = a^2$. Let $PQ$ and $PR$ be the tangents drawn from $P$ to the circle $x^2 + y^2 = a^2 \sin^2 \alpha$,where $Q$ and $R$ are the points of contact.Let $O$ be the origin $(0, 0)$. The radius of the inner circle is $OQ = a \sin \alpha$.The distance $OP$ is the radius of the outer circle,so $OP = a$.In the right-angled triangle $	riangle OQP$,$\sin(\angle OPQ) = \frac{OQ}{OP} = \frac{a \sin \alpha}{a} = \sin \alpha$.Therefore,$\angle OPQ = \alpha$.The angle between the tangents is $\angle QPR = 2 	imes \angle OPQ = 2\alpha$.

From any point on the circle $x^2 + y^2 = a^2$,tangents are drawn to the circle $x^2 + y^2 = a^2 \sin^2 \alpha$. The angle between them is:

Similar Questions

Suppose that the circle $x^2+y^2+2gx+2fy+c=0$ has its centre on $2x+3y-7=0$ and cuts the circles $x^2+y^2-4x-6y+11=0$ and $x^2+y^2-10x-4y+21=0$ orthogonally. Then $5g-10f+3c=$

The radical centre of the circles $x^2+y^2-4x-6y+5=0$,$x^2+y^2-2x-4y-1=0$ and $x^2+y^2-6x-2y=0$ is equal to

Where do the circles $x^2 + y^2 + 2x - 2y + 1 = 0$ and $x^2 + y^2 - 2x - 2y + 1 = 0$ touch each other?

If $(a, b)$ is the common point for the circles $x^2+y^2-4x+4y-1=0$ and $x^2+y^2+2x-4y+1=0$,then $a^2+b^2=$

The equation of the line joining the centres of the circles belonging to the coaxial system of circles $4x^2 + 4y^2 - 12x + 6y - 3 + \lambda(x + 2y - 6) = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module