Question

Find the equation of the circle which passes through the points of intersection of circles $x^2+y^2-2x-6y+6=0$ and $x^2+y^2+2x-6y+6=0$ and intersects the circle $x^2+y^2+4x+6y+4=0$ orthogonally.

• A. $x^2+y^2+14x-6y+6=0$
• B. $x^2+y^2+7x-3y+3=0$
• C. $x^2+y^2-14x+6y-3=0$
• D. $x^2+y^2+6x-4y+6=0$

Answer 1

Answer: (A)

$x^2+y^2+14x-6y+6=0$

The equation of the circle through the intersection of the given circles is

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

$\Rightarrow x^2+y^2-2x-6y+6+\lambda \left(4x\right)=0$

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

Where $4x=0$ is the equation of radical axis for the circle $x^2+y^2-2x-6y+6=0$ and $x^2+y^2+2x-6y+6=0$

It intersects at $x^2+y^2+4x+6y+4=0$ orthogonally

Hence $2g{g}_1+2f{f}_1=c+c_1$

where $g=-(2\lambda -1),f=3,g_1=-2,f_1=-3,c=6,c_1=4$

Now,$2g{g}_1+2f{f}_1=c+c_1$

$\Rightarrow 2\times -(2\lambda -1)\times -2+2\times 3\times -3=6+4$

$\Rightarrow 4(2\lambda -1)-18=10$

$\Rightarrow 4(2\lambda -1)=28$

$\Rightarrow 2\lambda -1=7$

$\Rightarrow 2\lambda =7+1=8$

$\therefore \lambda =4$

Hence the required equation of the circle is $x^2+y^2+2(2\lambda -1)x-6y+6=0$ at $\lambda =4$

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

or $x^2+y^2+14x-6y+6=0$