Question
The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length $$3a$$ is-
A.
$${x^2} + {y^2} = 9{a^2}$$
B.
$${x^2} + {y^2} = 16{a^2}$$
C.
$${x^2} + {y^2} = 4{a^2}$$
D.
$${x^2} + {y^2} = {a^2}$$
Answer :
$${x^2} + {y^2} = 4{a^2}$$
Solution :
Let $$ABC$$ be an equilateral triangle, whose median is $$AD.$$
Given $$AD = 3a$$
$$\eqalign{ & {\text{In }}\Delta ABD,\,A{B^2} = A{D^2} + B{D^2}; \cr & \Rightarrow {x^2} = 9{a^2} + \left( {\frac{{{x^2}}}{4}} \right)\,\,{\text{where}}\,\,AB = BC = AC = x \cr & \frac{3}{4}{x^2} = 9{a^2} \Rightarrow {x^2} = 12{a^2} \cr & {\text{In }}\Delta OBD,\,O{B^2} = O{D^2} + B{D^2}; \cr & \Rightarrow {r^2} = {\left( {3a - r} \right)^2} + \frac{{{x^2}}}{4} \cr & \Rightarrow {r^2} = 9{a^2} - 6ar + {r^2} + 3{a^2} \cr & \Rightarrow 6ar = 12{a^2} \cr & \Rightarrow r = 2a \cr} $$
So equation of circle is $${x^2} + {y^2} = 4{a^2}$$
Let $$ABC$$ be an equilateral triangle, whose median is $$AD.$$
Given $$AD = 3a$$
$$\eqalign{ & {\text{In }}\Delta ABD,\,A{B^2} = A{D^2} + B{D^2}; \cr & \Rightarrow {x^2} = 9{a^2} + \left( {\frac{{{x^2}}}{4}} \right)\,\,{\text{where}}\,\,AB = BC = AC = x \cr & \frac{3}{4}{x^2} = 9{a^2} \Rightarrow {x^2} = 12{a^2} \cr & {\text{In }}\Delta OBD,\,O{B^2} = O{D^2} + B{D^2}; \cr & \Rightarrow {r^2} = {\left( {3a - r} \right)^2} + \frac{{{x^2}}}{4} \cr & \Rightarrow {r^2} = 9{a^2} - 6ar + {r^2} + 3{a^2} \cr & \Rightarrow 6ar = 12{a^2} \cr & \Rightarrow r = 2a \cr} $$
So equation of circle is $${x^2} + {y^2} = 4{a^2}$$