Q. The equation of the circle passing through (1, 1) and the points of intersection of $x^{2} + y^{2} + 13x - 3y = 0$ and $2x^{2} + 2y^{2} + 4x - 7y - 25 = 0$ is
A) $4x^{2} + 4y^{2} - 30x - 10y = 25$
B) $4x^{2} + 4y^{2} + 30x - 13y - 25 = 0$
C) $4x^{2} + 4y^{2} - 17x - 10y + 25 = 0$
D) None of the above
Explanation
(b)
The required equation of circle is
$\left( x^{2} + y^{2} + 13x - 3y \right) + \lambda\left( 11x + \frac{1}{2}y + \frac{25}{2} \right) = 0$ ...(i)
It passes through (1, 1)
$\therefore 12 + \lambda(24) = 0$
$\Rightarrow \ \lambda = - \frac{1}{2}$
On putting in Eq. (i), we get
$x^{2} + y^{2} + 13x - 3y - \frac{11}{2}x - \frac{1}{4}y - \frac{25}{4} = 0$
$\Rightarrow \ 4x^{2} + 4y^{2} + 52x - 12y - 22x - y - 25 = 0$
$\Rightarrow \ 4x^{2} + 4y^{2} + 30x - 13y - 25 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.