Question

Equation of the circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length \(3\) is:

• A. \(x^2+y^2=4a^2\)
• B. \(x^2+y^2=2a^2\)
• C. \(x^2+y^2=9a^2\)
• D. \(x^2+y^2=16a^2\)

Hint:

Important formulas for an equilateral triangle of side \(s\): \[ \text{Median}=\frac{\sqrt3}{2}s \] \[ \text{Circumradius}=\frac{s}{\sqrt3} \] Circle centered at origin with radius \(R\): \[ x^2+y^2=R^2 \]

Answer 1

The Correct Option is A

Step 1: Understand the set up.
An equilateral triangle has all its vertices on one circle whose centre is the origin. The radius of that circle is the circumradius $R$. The median of the triangle is given as $3a$.
Step 2: Recall the median of an equilateral triangle.
For side $s$, the median (which is also the altitude) is $\dfrac{\sqrt3}{2}s$.
Step 3: Find the side from the median.
\[ \frac{\sqrt3}{2}s=3a\Rightarrow s=\frac{6a}{\sqrt3}=2\sqrt3\,a. \]
Step 4: Recall the circumradius formula.
For an equilateral triangle $R=\dfrac{s}{\sqrt3}$. The centroid and circumcentre are the same point, sitting two thirds along the median, which also gives $R=\dfrac{2}{3}\times\text{median}$.
Step 5: Compute the radius.
\[ R=\frac{2\sqrt3\,a}{\sqrt3}=2a. \] (Check: $\dfrac23\times3a=2a$, same answer.)
Step 6: Write the circle.
Centre at origin, radius $2a$, so $x^2+y^2=(2a)^2$. \[ \boxed{x^2+y^2=4a^2} \]