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Joint Entrance Examination
Mathematics
Circle
Question 170
Question 170 of 219
Easy
MCQ
The equation of a circle with origin as a center and passing through an equilateral triangle whose median is of length
3
3
3
a
a
a
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
\[{x^2}\, + \,{y^2} = {a^2}\]
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Official solution
Let
A
B
C
ABC
A
B
C
be an equilateral triangle, whose median is
A
D
.
AD.
A
D
.
Given
A
D
=
3
a
.
AD=3a.
A
D
=
3
a
.
In \[\Delta ABD,\,\,A{B^2} = A{D^2} + B{D^2};\] \[ \Rightarrow {x^2} = 9{a^2} + \left( {{x^2}/4} \right)\,\,\] where
A
B
=
B
C
=
A
C
=
x
.
AB = BC = AC = x.
A
B
=
B
C
=
A
C
=
x
.
\[{3 \over 4}{x^2} = 9{a^2} \Rightarrow {x^2} = 12{a^2}.\] In \[\,\,\,\Delta OBD,O{B^2} = O{D^2} + B{D^2}\] \[ \Rightarrow {r^2} = {\left( {3a - r} \right)^2} + {{{x^2}} \over 4}\] \[ \Rightarrow {r^2} = 9{a^2} - 6ar + {r^2} + 3{a^2};\] \[ \Rightarrow 6ar = 12{a^2}\] \[ \Rightarrow r = 2a\] So equation of circle is
x
2
+
y
2
=
4
a
2
{x^2} + {y^2} = 4{a^2}
x
2
+
y
2
=
4
a
2
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