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

Question

The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is (A) x2+y2=9a2 (B) x2+y2=16a2 (C) x2+y2=4a2 (D) x2+y2=a2

Tardigrade Admin · Accepted Answer

We know that centroid divides the median in the ratio

2 : 1

Radius of the circle

= \frac{2}{3} \times

length of median

= \frac{2}{3} \times 3 a = 2 a

Centre of the (given) circle is

C (0, 0)

Therefore the equation of the circle

(x - 0)^{2} + (y - 0)^{2} = (2 a)^{2} \Rightarrow x^{2} + y^{2} = 4 a^{2}

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

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

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

$x^{2} + y^{2} = 4 a^{2}$

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

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

x2+y2=9a2

x2+y2=16a2

x2+y2=4a2

x2+y2=a2

We know that centroid divides the median in the ratio 2:1 Radius of the circle =32​× length of median =32​×3a=2a Centre of the (given) circle is C(0,0) Therefore the equation of the circle (x−0)2+(y−0)2=(2a)2⇒x2+y2=4a2

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

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

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

$x^{2} + y^{2} = 4 a^{2}$

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