Q. The centre of the circle $r^{2} - 4r\left( \cos{\theta + \sin\theta} \right) - 4 = 0$ in Cartesian coordinates is
A) (1, 1)
B) $( - 1,\ - 1)$
C) (2, 2)
D) $( - 2,\ - 2)$
Explanation
(c)
Given, $r^{2} - 4r\left( \cos{\theta + \sin\theta} \right) - 4 = 0$ …(i)
Put $x = r\cos\theta,\ \ y = r\sin\theta$, then $r^{2} = x^{2} + y^{2}$
$\therefore$ From Eq. (i)
$x^{2} + y^{2} - 4(x + y) - 4 = 0$
$\Rightarrow \ x^{2} + y^{2} - 4x - 4y - 4 = 0$
$\therefore$ Centre of circle is (2, 2)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.