Let AB be a line segment with mid-point C and | vedclass

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

Question

(A) Let the radius of circle $C_1$ be $2a$. Place the center $C$ at the origin $(0,0)$ and $AB$ along the $X$-axis. Then $B = (2a, 0)$,$A = (-2a, 0)$,

Vedclass · Accepted Answer

$\frac{1}{\sqrt{10}}$

Vedclass · Answer

(A) Let the radius of circle $C_1$ be $2a$. Place the center $C$ at the origin $(0,0)$ and $AB$ along the $X$-axis. Then $B = (2a, 0)$,$A = (-2a, 0)$,$C = (0, 0)$,and $D = (-a, 0)$.Since $E$ is on $C_1$ and $EC \perp AB$,$E$ has coordinates $(0, 2a)$.Since $F$ is on $C_2$ (diameter $AC$,center $D(-a, 0)$,radius $a$) and $DF \perp AB$,$F$ has coordinates $(-a, -a)$.We need to find $\sin \angle FEC$. Let $\angle FEC = 	heta$.Vector $\vec{EC} = C - E = (0, 0) - (0, 2a) = (0, -2a)$.Vector $\vec{EF} = F - E = (-a, -a) - (0, 2a) = (-a, -3a)$.The angle $	heta$ between $\vec{EC}$ and $\vec{EF}$ is given by $\cos 	heta = \frac{\vec{EC} \cdot \vec{EF}}{|\vec{EC}| |\vec{EF}|}$.$\vec{EC} \cdot \vec{EF} = (0)(-a) + (-2a)(-3a) = 6a^2$.$|\vec{EC}| = \sqrt{0^2 + (-2a)^2} = 2a$.$|\vec{EF}| = \sqrt{(-a)^2 + (-3a)^2} = \sqrt{a^2 + 9a^2} = a\sqrt{10}$.$\cos 	heta = \frac{6a^2}{(2a)(a\sqrt{10})} = \frac{6}{2\sqrt{10}} = \frac{3}{\sqrt{10}}$.Since $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{9}{10} = \frac{1}{10}$,we have $\sin 	heta = \frac{1}{\sqrt{10}}$.

Similar Questions

Number of common tangents to the circles $x^2+y^2-6x-14y+48=0$ and $x^2+y^2-6x=0$ are

The radius of any circle touching the lines $3x - 4y + 5 = 0$ and $6x - 8y - 9 = 0$ is

If the straight line $y=mx$ lies outside of the circle $x^2+y^2-20y+90=0$, then the value of $m$ will satisfy

In the circle given below,let $OA = 1$ unit,$OB = 13$ units and $PQ \perp OB$. Then,the area of the triangle $PQB$ (in square units) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module