Q. The equation of a circle with origin as a centre and passing through an equilateral triangle whose median is of length $3a$, is
A) $x^{2} + y^{2} = 9a^{2}$
B) $x^{2} + y^{2} = 16a^{2}$
C) $x^{2} + y^{2} = 4a^{2}$
D) $x^{2} + y^{2} = a^{2}$
Explanation
(c)
In an equilateral triangle, the circumcentre of a circle lies on the centroid of the triangle
Here, radius of circle is $2a$
[image:11_image89.emf]
$\therefore$ Required equation of circle is
$x^{2} + y^{2} = 4a^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.