Question

Find the equation of family of circles through the intersection of $x^2+y^2-6x+2y+4=0$ and $x^2+y^2+2x-4y-6=0$ whose center lies on $y=x.$

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Given circles

$S_1=0x^2+y^2-6x+2y+4=0$ - - - - - - -(1)

$S_2=0x^2+y^2-2x-4y-6=0$ - - - - - - -(2)

The equation of family of circles passing through the points of intersection of two circles $S_1=0\&S_2=0$ is $S_1+\lambda S_2=0$

$(x^2+y^2-6x+2y+4)+\lambda (x^2+y^2+2x-4y-6)=0$ - - - - - - - (3)

$(1+\lambda )x^2+(1+\lambda )y^2+(2\lambda -6)x+(2-4\lambda )y+4-6\lambda =0$

Dividing both sides by $1+\lambda$

$x^2+y^2+\frac{2(\lambda -3)}{1+\lambda }x+\frac{2(1-2\lambda )}{1+\lambda }y+\frac{4-6\lambda }{1+\lambda }=0$

Coordinates of center of circle is (-g,-f)

$(\frac{3-\lambda }{1+\lambda },\frac{2\lambda -1}{1+\lambda })$

Since, center lies on the line $y=x$

$\frac{3-\lambda }{1+\lambda }=\frac{2\lambda -1}{1+\lambda }$

$\left\{\lambda \ne -1\right\}$

$\lambda =\frac{4}{3}$

Equation of family of circles passing through the point of intersection of circle is

$(x^2+y^2-6x+2y+4)+\frac{4}{3}(x^2+y^2+2x-4y-6)=0$

$\frac{7}{3}{x}^2+\frac{7}{3}{y}^2-\frac{10}{3}x-\frac{10}{3}y-4=0$

$x^2+y^2-\frac{10}{7}x-\frac{10}{7}y-\frac{12}{7}=0$

Option A is correct