The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length $3a$ is (A) $x^2 + y^2 = 9a^2$ (B) $x^2 + y^2 = 16a^2$ (C) $x^2 + y^2 = 4a^2$ (D) $x^2 + y^2 = a^2$

The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle who

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Concept understanding — Circle Equation Standard Form

Where the Circle Equation Comes From

Imagine you're standing at a point on a flat field. You tie a rope to a stake at that point, walk out the full length of the rope, and start walking in a circle, keeping the rope taut. Every point you step on is exactly the same distance from the stake.

That's the entire idea: a circle is the set of all points that are a fixed distance (the radius) from a fixed point (the centre).

If we put this on a coordinate plane, we can turn that geometric idea into an algebraic equation.

From Geometry to Algebra

Let the centre be at coordinates $(h, k)$ . Let the radius be $r$ . Take any point $(x, y)$ that lies on the circle. The distance from $(x, y)$ to $(h, k)$ must equal $r$ .

What's the distance between two points in the plane? The distance formula:

\sqrt{(x - h)^2 + (y - k)^2} = r

Now square both sides to remove the square root:

(x - h)^2 + (y - k)^2 = r^2

That's it. That's the standard form of the equation of a circle.

(x - h)^2 + (y - k)^2 = r^2

What Each Piece Tells You

$(x - h)$ and $(y - k)$ — these shift the circle away from the origin. If the centre is at $(0,0)$ , the equation simplifies to $x^2 + y^2 = r^2$ .
$r^2$ — notice it's the square of the radius, not the radius itself. If the equation says $x^2 + y^2 = 25$ , the radius is $\sqrt{25} = 5$ , not 25.
The equals sign — only the points $(x, y)$ that make this equation true lie on the circle. Any other point gives a larger or smaller left-hand side.

Watch out

A common mistake: for $(x - 3)^2 + (y + 2)^2 = 16$ , students often read the centre straight off the signs printed in the equation and say $(3, 2)$ . That's wrong. Each bracket must first be written in the exact form $x - h$ and $y - k$ : here $(y + 2) = (y - (-2))$ , so $k = -2$ , not $2$ . The centre is actually $(3, -2)$ . Always flip the sign inside every bracket before reading off $h$ and $k$ .

Quick Example

Write the equation of a circle with centre $(-1, 4)$ and radius $3$ .

Here $h = -1$ , $k = 4$ , $r = 3$ . Plug in:

(x - (-1))^2 + (y - 4)^2 = 3^2

Simplify:

$(x + 1)^2 + (y - 4)^2 = 9$ …

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The key idea is that the circumcenter of an equilateral triangle coincides with its centroid, and the circumradius equals the distance from the centroid to any vertex. Given the median length $3a$ , the centroid divides it in the ratio $2:1$ , so the circumradius is $\frac{2}{3} \times 3a = 2a$ , giving the circle equation $x^2 + y^2 = 4a^2$ . The correct option is (C).

We start with the geometry of an equilateral triangle. In any triangle, the centroid, circumcenter, and orthocenter are distinct points — but in an equilateral triangle, all three coincide. This is a powerful simplification: the center of the circumscribed circle (circumcenter) is exactly the centroid.

The centroid is the intersection of the medians. A median of a triangle is a line from a vertex to the midpoint of the opposite side. For an equilateral triangle, all medians are equal in length, and each median is also an altitude and an angle bisector.

Given: median length $= 3a$ .

The centroid divides each median in the ratio $2:1$ , with the longer segment from the vertex to the centroid. So the distance from the centroid (which is also the circumcenter) to any vertex is $\frac{2}{3}$ of the median length.

Find the circumradius The circumradius $R$ is the distance from the circumcenter to any vertex. Since the centroid is the circumcenter:

R = \frac{2}{3} \times (\text{median length}) = \frac{2}{3} \times 3a = 2a.

Equation of the circle The circle has its center at the origin $(0,0)$ and radius $R = 2a$ . …

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