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Circle question

2002 · Shift 0 · Q83

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JEE Main
Mathematics
Circle
2002 · Shift 0 · Q83

Circle question

2002 · Shift 0 · Q83

JEE MainMathematicsCircleMCQ+4 / −1

The equation of a circle with origin as a center and passing through an 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
${x^2}\, + \,{y^2} = {a^2}$

View written solutionFree

Correct answer: C

Interpretation of the question

A circle has center at the origin, and it is associated with an equilateral triangle whose median is $3a$ .

For an equilateral triangle, all medians, altitudes, angle bisectors, and perpendicular bisectors coincide at the same center.

If the circle is the one passing through the vertices of the equilateral triangle, then it is the circumcircle of the triangle.

So we need the circumradius of the equilateral triangle.

Use the median of the equilateral triangle

Let the side of the equilateral triangle be $s$ .

In an equilateral triangle,

\text{median} = \text{altitude} = \frac{\sqrt{3}}{2}s

Given median $= 3a$ , so

\frac{\sqrt{3}}{2}s = 3a

Hence,

s = \frac{6a}{\sqrt{3}} = 2\sqrt{3}a

Find the circumradius

For an equilateral triangle of side $s$ , the circumradius is

R = \frac{s}{\sqrt{3}}

Substituting $s = 2\sqrt{3}a$ ,

R = \frac{2\sqrt{3}a}{\sqrt{3}} = 2a

So the required circle has radius $2a$ .

Equation of the circle

A circle centered at the origin with radius $R$ has equation

x^2 + y^2 = R^2

Thus,

x^2 + y^2 = (2a)^2 = 4a^2

Check the options

A: $x^2+y^2=9a^2$ ❌
B: $x^2+y^2=16a^2$ ❌
C: $x^2+y^2=4a^2$ ✅
D: $x^2+y^2=a^2$ ❌

Therefore, the correct option is C.

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