Suppose A(2,3) and B are the points of inters | vedclass

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

Question

(B) Let $C_1$ and $C_2$ be the centers of the two circles. Since $BP$ and $BQ$ are diameters,$C_1$ is the midpoint of $BP$ and $C_2$ is the midpoint o

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Let $C_1$ and $C_2$ be the centers of the two circles. Since $BP$ and $BQ$ are diameters,$C_1$ is the midpoint of $BP$ and $C_2$ is the midpoint of $BQ$.In $	riangle BPQ$,$C_1$ and $C_2$ are the midpoints of sides $BP$ and $BQ$ respectively. By the midpoint theorem,$PQ \parallel C_1C_2$.The line joining the centers $C_1C_2$ is perpendicular to the radical axis $AB$.Therefore,$PQ$ is perpendicular to the radical axis $AB$.If the slope of the radical axis is $m_1 = 3/4$ and the slope of $PQ$ is $m_2 = a/b$,then $m_1 	imes m_2 = -1$.$\frac{3}{4} 	imes \frac{a}{b} = -1$$\frac{3a}{4b} = -1$$3a = -4b$$3a + 4b = 0$.

Similar Questions

The line $x-2=0$ cuts the circle $x^2+y^2-8x-2y+8=0$ at $A$ and $B$. The equation of the circle passing through the points $A$ and $B$ and having the least radius is

The equation of the circle which passes through the point $(3,2)$,bisects the circumference of the circle $x^2+y^2=15$,and cuts the circle $x^2+y^2+4x+6y+3=0$ orthogonally is

If the equation of the common tangent of the circles $x^2+y^2-4x+6y+4=0$ and $x^2+y^2+2x-2y-2=0$ at their point of contact is $ax+by+c=0$,then $\frac{a}{c}=$

If the angle between the circles $x^2+y^2-2x+2y+1=0$ and $x^2+y^2+2x-2y+k=0$ is $\frac{\pi}{3}$,then

If the acute angle between the circles $S \equiv x^2+y^2+2kx+4y-3=0$ and $S' \equiv x^2+y^2-4x+2ky+9=0$ is $\cos^{-1}(\frac{3}{8})$ and the centre of $S'=0$ lies in the first quadrant,then the radical axis of $S=0$ and $S'=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module