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# The equation of circle with origin as a centre and passing though 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) ${x^2} + {y^2} = {a^2}$

Last updated date: 18th Jun 2024
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Hint: Equation of circle having radius (r) and passing through point$\left( {{x_1},{y_1}} \right)$ is given as : $\Rightarrow {(x - {x_1})^2} + {(y - {y_1})^2} = {(radius)^2}$.

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side, thus bisecting that side.

Let

The length of the side of the triangle be x cm and the radius of the circle be r cm.

Since, it is an equilateral triangle

$\Rightarrow$AB=BC=AC=x cm.

Now, In $\vartriangle ABD$,

AD is a median and a median in an equilateral triangle will always be perpendicular.

$BD = \dfrac{1}{2}AB$( $\because$AD median bisects BC).

$\Rightarrow {\left( {AB} \right)^2} = {(AD)^2} + {(BD)^2}$

$\Rightarrow {x^2} = {\left( {3a} \right)^2} + {\left( {\dfrac{x}{2}} \right)^2}$

$\Rightarrow {x^2} = 9{a^2} + {\left( {\dfrac{x}{2}} \right)^2}$

$\Rightarrow {x^2} - \dfrac{{{x^2}}}{4} = 9{a^2}$

$\Rightarrow \dfrac{3}{4}{x^2} = 9{a^2}$

$\Rightarrow {x^2} = 12{a^2}$

$\Rightarrow$${\left( {AB} \right)^2}$=$12{a^2}$

Since, ${\left( {BD} \right)^2} = \dfrac{1}{4}{\left( {AB} \right)^2}$

$\Rightarrow {\left( {BD} \right)^2} = \dfrac{1}{4}{(x)^2}$

$\Rightarrow {\left( {BD} \right)^2} = \dfrac{1}{4}(12{a^2})$

$\Rightarrow {\left( {BD} \right)^2} = 3{a^2}$

Now In $\vartriangle OBD$,

$\Rightarrow {\left( {OB} \right)^2} = {(BD)^2} + {(OD)^2}$

$\Rightarrow {r^2} = {\left( {3a - r} \right)^2} + 3{a^2}......(\because OB = radius = rcm)$

$\Rightarrow {r^2} = 9{a^2} + {r^2} - 6ar + 3{a^2}$

$\Rightarrow 12{a^2} = 6ar$

$\Rightarrow r = 2a$

Equation of circle having radius (r) and passing through point$\left( {{x_1},{y_1}} \right)$ is given as : $\Rightarrow {(x - {x_1})^2} + {(y - {y_1})^2} = {(radius)^2}$

As given circle passes through origin i.e. $O(0,0)$

$\Rightarrow {(x - 0)^2} + {(y - 0)^2} = {(r)^2}$

$\Rightarrow {(x - 0)^2} + {(y - 0)^2} = {(2a)^2}......(\because r = 2a)$

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

Required equation of circle is given as: ${x^2} + {y^2} = 4{a^2}$

Option (C) is correct.

Note: Equilateral triangles have all sides equal.

A median in an equilateral triangle will always be perpendicular to the opposite side of the triangle.

Always make diagrams while solving such questions.