The equation of a circle with origin as a centre and passing through equilateral triangle whose medi

Q: Wait, why is the origin the centroid? I thought the centroid was just the average of the vertices.

You are absolutely right! The centroid is the average of the vertices. In an equilateral triangle, the symmetry is so perfect that the circumcenter, the centroid, and the orthocenter all collapse into the same point. Since the circle is centered at the origin and passes through the vertices, the origin is the circumcenter, and therefore, it must also be the centroid.

Q: Why does the centroid divide the median in a $2:1$ ratio? Is this always true?

Yes, this is a fundamental property of all triangles! The centroid is the center of mass, and it always divides each median into two segments with a ratio of $2:1$, where the segment connecting the vertex to the centroid is twice as long as the segment connecting the centroid to the midpoint of the opposite side. This is a powerful tool in your geometry toolkit.

Q: Can I solve this by finding the coordinates of the vertices?

You certainly could! You could place the vertices at $(0, h)$, $(-\frac{s}{2}, -\frac{h}{3})$, and $(\frac{s}{2}, -\frac{h}{3})$ where $h$ is the height (median) and $s$ is the side length. However, that path is much longer and prone to algebraic errors. Using the geometric properties of the centroid is the elegant, 'JEE-style' way to solve this quickly and accurately.

Question

The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length $3a$ is

Options:

(A) $x^2 + y^2 = 9a^2$

(B) $x^2 + y^2 = 16a^2$

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

(D) $x^2 + y^2 = a^2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at a circle perfectly centered at the origin $(0, 0)$. This circle is a guardian, passing through the three vertices of an equilateral triangle.

This setup is a classic problem that tests your ability to bridge the gap between coordinate geometry and pure Euclidean geometry.

### The Centroid-Circumcenter Coincidence

The first step is to visualize the symmetry. We know the circle is centered at the origin, which means the origin is the circumcenter of the triangle.

In an equilateral triangle, the centroid, the circumcenter, and the orthocenter are all the same point. This means the origin $(0, 0)$ is also the centroid of our triangle.

This realization is the key that unlocks the entire problem.

### The Median Mystery

We are given that the median of this triangle has a length of $3a$. Recall the fundamental property of the centroid: it divides every median in a $2:1$ ratio.

The segment from the vertex to the centroid is twice as long as the segment from the centroid to the midpoint of the opposite side. In our case, the distance from the vertex to the centroid is exactly the circumradius $R$ of the circle.

Therefore, we can write the relationship as:

$$R = \frac{2}{3} 	imes 	ext{Median}$$

### The Calculation

With the formula $R = \frac{2}{3} 	imes 	ext{Median}$ in our toolkit, we simply substitute the given length of the median, $3a$.

$$R = \frac{2}{3} 	imes (3a)$$

The threes cancel out with elegant simplicity, leaving us with $R = 2a$. We have successfully determined the radius of the circle.

### The Final Equation

Now, we return to the standard equation of a circle centered at the origin: $x^2 + y^2 = R^2$. Substituting our value of $R = 2a$, we get:

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

Squaring the term $2a$ gives us $4a^2$. Thus, the final equation of our circle is:

**$$x^2 + y^2 = 4a^2$$**

It is a beautiful, concise result that emerges from understanding the deep geometric soul of the triangle. Geometry is not just about formulas; it is about seeing the hidden connections.

Animated Solution for Mathematics - Circles: The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length $3 a$ is

Select Answer:

Visualized Solution

Visualizing the Coordinate System

The Equilateral Triangle

Drawing the Median

Centroid and Circumcenter

Locating the Centroid

The Centroid Ratio Property

Formula for Circumradius

Substituting the Median Length

Calculating the Radius $R$

Standard Circle Equation

Substituting the Radius

Expanding the Equation

Final Conclusion

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Centroid-Circumcenter Coincidence

The Median Mystery

The Calculation

The Final Equation

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is