Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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) \[{x^2} + {y^2} = {a^2}\]

seo-qna
Last updated date: 18th Jun 2024
Total views: 403.2k
Views today: 9.03k
Answer
VerifiedVerified
403.2k+ views

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.


Complete step-by-step answer:

Let

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

seo images

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.