The equation of a circle passing through the | vedclass

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

Question

(A) The equation of a circle passing through the points of intersection of the two circles is given by $S_1 + \lambda S_2 = 0$,where $\lambda 
eq -1$

Vedclass · Accepted Answer

$x^2+y^2-2x-12=0$

Vedclass · Answer

(A) The equation of a circle passing through the points of intersection of the two circles is given by $S_1 + \lambda S_2 = 0$,where $\lambda 
eq -1$. $(x^2+y^2-4x-6y-12) + \lambda(x^2+y^2+6x+4y-12) = 0$ $(1+\lambda)x^2 + (1+\lambda)y^2 + (6\lambda-4)x + (4\lambda-6)y - 12(1+\lambda) = 0$ Dividing by $(1+\lambda)$,we get $x^2+y^2 + \frac{6\lambda-4}{1+\lambda}x + \frac{4\lambda-6}{1+\lambda}y - 12 = 0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we have $g = \frac{3\lambda-2}{1+\lambda}$,$f = \frac{2\lambda-3}{1+\lambda}$,and $c = -12$. The radius $r = \sqrt{g^2+f^2-c} = \sqrt{13}$. $g^2+f^2-c = 13 \implies \frac{(3\lambda-2)^2 + (2\lambda-3)^2}{(1+\lambda)^2} + 12 = 13$. $(3\lambda-2)^2 + (2\lambda-3)^2 = (1+\lambda)^2$. $9\lambda^2 - 12\lambda + 4 + 4\lambda^2 - 12\lambda + 9 = 1 + 2\lambda + \lambda^2$. $12\lambda^2 - 26\lambda + 12 = 0 \implies 6\lambda^2 - 13\lambda + 6 = 0$. $(2\lambda-3)(3\lambda-2) = 0$,so $\lambda = \frac{3}{2}$ or $\lambda = \frac{2}{3}$. For $\lambda = \frac{2}{3}$,the equation becomes $x^2+y^2-2x-12=0$. For $\lambda = \frac{3}{2}$,the equation becomes $x^2+y^2+2x-12=0$.

List-$I$	List-$II$
$A$. $x^2+y^2+2x+8y-23=0$,$x^2+y^2-4x-10y+19=0$	$I$. $0$
$B$. $x^2+y^2=1$,$x^2+y^2-2x-6y+6=0$	$II$. $1$
$C$. $x^2+y^2-8x+2y=0$,$x^2+y^2-2x-16y+25=0$	$III$. $2$
$D$. $x^2+y^2=4$,$x^2+y^2-2x=0$	$IV$. $3$
	$V$. $4$

The equation of a circle passing through the points of intersection of the circles $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+4y-12=0$ and having radius $\sqrt{13}$ is

Similar Questions

If the circles $x^2 + y^2 + 2ax + c = 0$ and $x^2 + y^2 + 2by + c = 0$ touch each other,then:

If $y = 2x$ is a chord of the circle $x^{2} + y^{2} = 10x$,then find the equation of the circle having this chord as its diameter.

The point $(3, -4)$ lies on both the circles $x^2 + y^2 - 2x + 8y + 13 = 0$ and $x^2 + y^2 - 4x + 6y + 11 = 0$. Then,the angle between the circles is

If $P$ is the point of contact of the circles $x^2+y^2+4x+4y-10=0$ and $x^2+y^2-6x-6y+10=0$,and $Q$ is their external centre of similitude,then the equation of the circle with $P$ and $Q$ as the extremities of its diameter is

Vedclass Test Series

Exam Paper Generator

Online Exam Module