The length of the common chord of the circles | vedclass

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

Question

(A) Given circles are:$S_1: x^2+y^2-6x-4y+13-c^2=0$$S_2: x^2+y^2-4x-6y+13-c^2=0$Centres are $C_1(3,2)$ and $C_2(2,3)$ with radius $r_1=r_2=c$.The equa

Vedclass · Accepted Answer

$\sqrt{4c^2-2}$

Vedclass · Answer

(A) Given circles are:$S_1: x^2+y^2-6x-4y+13-c^2=0$$S_2: x^2+y^2-4x-6y+13-c^2=0$Centres are $C_1(3,2)$ and $C_2(2,3)$ with radius $r_1=r_2=c$.The equation of the common chord is given by $S_1-S_2=0$:$(x^2+y^2-6x-4y+13-c^2)-(x^2+y^2-4x-6y+13-c^2)=0$$-2x+2y=0 \Rightarrow x-y=0$.Let $M$ be the midpoint of the common chord $AB$. Then $C_1M$ is the perpendicular distance from $C_1(3,2)$ to the line $x-y=0$:$C_1M = \frac{|3-2|}{\sqrt{1^2+(-1)^2}} = \frac{1}{\sqrt{2}}$.In $	riangle AC_1M$,$AM^2 = AC_1^2 - C_1M^2 = c^2 - (\frac{1}{\sqrt{2}})^2 = c^2 - \frac{1}{2}$.$AM = \sqrt{c^2 - \frac{1}{2}} = \sqrt{\frac{2c^2-1}{2}} = \frac{\sqrt{4c^2-2}}{2}$.The length of the common chord $AB = 2AM = 2 	imes \frac{\sqrt{4c^2-2}}{2} = \sqrt{4c^2-2}$.

The length of the common chord of the circles $x^2+y^2-6x-4y+13-c^2=0$ and $x^2+y^2-4x-6y+13-c^2=0$ is

Similar Questions

Two circles whose radii are equal to $4$ and $8$ intersect at right angles. The length of their common chord is:

Let $L_1$ be the length of the common chord of the curves $x^2 + y^2 = 9$ and $y^2 = 8x$,and $L_2$ be the length of the latus rectum of $y^2 = 8x$,then

The length of the common chord of the two circles $x^2+y^2-4x-8y+4=0$ and $x^2+y^2-8x-12y+16=0$ is

If the circles $x^{2}+y^{2}-2x-2y-7=0$ and $x^{2}+y^{2}+4x+2y+k=0$ cut orthogonally,then the length of the common chord of the circles is

If tangents are drawn to the circle $x^2+y^2=12$ at the points of intersection with the circle $x^2+y^2-5x+3y-2=0$,then the ordinate of the point of intersection of these tangents is

Vedclass Test Series

Exam Paper Generator

Online Exam Module