Q. The equation of the circle passing through (4, 5) and having the centre (2, 2), is
A) $x^{2} + y^{2} + 4x + 4y - 5 = 0$
B) $x^{2} + y^{2} - 4x - 4y - 5 = 0$
C) $x^{2} + y^{2} - 4x$=13
D) $x^{2} + y^{2} - 4x - 4y + 5 = 0$
Explanation
(b)
Equation of circle whose centre is at (2, 2) and radius $r$ is
$(x - 2)^{2} + (y - 2)^{2} = r^{2}$ …(i)
This circle passes through (4, 5), then
$(4 - 2)^{2} + (5 - 2)^{2} = r^{2}$
$\Rightarrow \ \ \ \ r^{2} = 13$
On putting this values in Eq. (i), we get
$(x - 2)^{2} + (y - 2)^{2} - 13 = 0$
$\Rightarrow \ x^{2} + y^{2} - 4x - 4y - 5 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.