Q. The equation of the circle which passes through the origin and cuts orthogonally each of the circles $x^{2} + y^{2} - 6x + 8 = 0$ and $x^{2} + y^{2} - 2x - 2y - 7 = 0$ is
A) $3x^{2} + 3y^{2} - 8x - 13y = 0$
B) $3x^{2} + 3y^{2} - 8x + 29y = 0$
C) $3x^{2} + 3y^{2} + 8x + 29y = 0$
D) $3x^{2} + 3y^{2} - 8x - 29y = 0$
Explanation
(b)
Let the required equation of circle be$x^{2} + y^{2} + 2gx + 2fy = 0$. Since, the above circle cuts the given circles orthogonally
$\therefore\ \ 2( - 3g) + 2f(0) = 8\ \ \ \Rightarrow \ \ 2g = - \frac{8}{3}$
And $- 2g - 2f = - 7$
$\Rightarrow \ 2f = - 7 + \frac{8}{3} = \frac{29}{3}$
$\therefore$ required equation of circle is
$x^{2} + y^{2} - \frac{8}{3}x + \frac{29}{3}y = 0$
or $3x^{2} + 3y^{2} - 8x + 29y = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.