Question:
<p>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</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> = 9a<sup>2</sup></p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> = 4a<sup>2</sup></p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> = 16a<sup>2</sup></p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> = a<sup>2</sup></p>
Step-by-Step Solution
Key Concept: For an equilateral triangle, the circumcenter and centroid coincide, and the circumradius is calculated as two-thirds of the median length.
<p>Since the triangle is equilateral.<br />
<span class="math-tex">$\therefore$</span> The centroid of the triangle is the same as the circumcentre.<br />
Radius of the circumcircle = <span class="math-tex">$\frac{2}{3}$</span> (median) = <span class="math-tex">$\frac{2}{3}$</span>(3a) = 2a<br />
Hence, the equation of the circumcircle whose centre is at (0, 0) and radius 2a is x<sup>2</sup> + y<sup>2</sup> = (2a)<sup>2</sup>.</p>
Correct Answer: B