Q. The centre of a circle is $(2,\ - 3)$ and the circumference is$10\pi$. Then, the equation of the circle is
A) $x^{2} + y^{2} + 4x + 6y + 12 = 0$
B) $x^{2} + y^{2} - 4x + 6y + 12 = 0$
C) $x^{2} + y^{2} - 4x + 6y - 12 = 0$
D) $x^{2} + y^{2} - 4x - 6y - 12 = 0$
Explanation
(c)
It is given, centre is $(2,\ - 3)$ and circumference of circle $= 10\pi$
$\Rightarrow \ \ 2\pi r = 10\pi\ \Rightarrow r = 5$
The equation of circle, if centre is $(2,\ - 3)$ and radius is 5, is
$(x - 2)^{2} + (y + 3)^{2} = 5^{2}$
$\Rightarrow \ \ x^{2} + y^{2} - 4x + 6y + 13 = 25$
$\Rightarrow x^{2} + y^{2} - 4x + 6y - 12 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.