Question

A($a$, 0) is a fixed point, and $\theta$ is a parameter such that $0<\theta<2\pi$. If P($a \cos \theta$, $a \sin \theta$) is a point on the circle $x^2 + y^2 = a^2$ and Q($b \sin \theta$, $-b \cos \theta$) is a point on the circle $x^2 + y^2 = b^2$, then the locus of the centroid of the triangle APQ is

• A. a circle with centre at $\left( \frac{a}{3}, 0 \right)$ and radius $\frac{\sqrt{a^2 + b^2}}{3}$
• B. a circle with centre at $(a, 0)$ and radius $\frac{\sqrt{a^2 + b^2}}{3}$
• C. a parabola with focus at $\left( \frac{a}{3}, 0 \right)$
• D. a parabola with focus at $(a, 0)$

Hint:

To find the locus of a centroid, compute the centroid’s coordinates and eliminate the parameter by squaring and adding to form a conic equation.

Answer 1

The Correct Option is A

Given points $A(a, 0)$, $P(a \cos \theta, a \sin \theta)$, and $Q(b \sin \theta, -b \cos \theta)$, the centroid $G(h, k)$ of triangle $APQ$ is: \[ h = \frac{a + a \cos \theta + b \sin \theta}{3}, \quad k = \frac{0 + a \sin \theta - b \cos \theta}{3} \] Rewrite: \[ 3h - a = a \cos \theta + b \sin \theta, \quad 3k = a \sin \theta - b \cos \theta \] Square and add to eliminate $\theta$: \[ (3h - a)^2 + (3k)^2 = (a \cos \theta + b \sin \theta)^2 + (a \sin \theta - b \cos \theta)^2 \] \[ = a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \sin \theta \cos \theta + a^2 \sin^2 \theta + b^2 \cos^2 \theta - 2ab \sin \theta \cos \theta \] \[ = a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) = a^2 + b^2 \] \[ \left( h - \frac{a}{3} \right)^2 + k^2 = \frac{a^2 + b^2}{9} \] This is a circle with centre $\left( \frac{a}{3}, 0 \right)$ and radius $\frac{\sqrt{a^2 + b^2}}{3}$. Option (1) is correct. Options (2), (3), and (4) do not describe the derived locus.