The equation of the circle with origin as centre passing the vertices of an equilateral triangle whose median is of length 3a is

Question

The equation of the circle with origin as centre passing the vertices of an equilateral triangle whose median is of length 3a is (A) x2 +y2 =9a2 (B) x2 +y2 =16a2 (C) x2 +y2 =a2 (D) None of the above

Tardigrade Admin · Accepted Answer

Centre (0,0), radius =3 a × (2/3)=2 a Hence, circle x2+y2=4 a2 as centroid divides median in the ratio of 2: 1.

Q. The equation of the circle with origin as centre passing the vertices of an equilateral triangle whose median is of length $3 a$ is

$x^{2} + y^{2} = 9 a^{2}$

$x^{2} + y^{2} = 16 a^{2}$

$x^{2} + y^{2} = a^{2}$

None of the above

Centre $(0, 0)$ , radius $= 3 a \times \frac{2}{3} = 2 a$
Hence, circle $x^{2} + y^{2} = 4 a^{2}$ as centroid divides
median in the ratio of $2 : 1$ .

Q. The equation of the circle with origin as centre passing the vertices of an equilateral triangle whose median is of length 3a is

x2+y2=9a2

x2+y2=16a2

x2+y2=a2