Question

Let a circle of radius $4$ pass through the origin $O$, the points $A(-\sqrt{3}a,0)$ and $B(0,-\sqrt{2}b)$, where $a$ and $b$ are real parameters and $ab\neq0$. Then the locus of the centroid of $\triangle OAB$ is a circle of radius

• A. $\dfrac{8}{3}$
• B. $\dfrac{5}{3}$
• C. $\dfrac{11}{3}$
• D. $\dfrac{7}{3}$

Hint:

When dealing with centroids, always express parameters in terms of centroid coordinates to obtain the locus.

Answer 1

The Correct Option is A

Step 1: Coordinates of the centroid.
Centroid $G$ of $\triangle OAB$ is \[ G\left(\frac{-\sqrt{3}a}{3},\frac{-\sqrt{2}b}{3}\right) \] Step 2: Using the circle condition.
Since the circle of radius $4$ passes through $O,A,B$, \[ OA^2+OB^2=16 \] \[ (\sqrt{3}a)^2+(\sqrt{2}b)^2=16 \Rightarrow 3a^2+2b^2=16 \] Step 3: Writing locus equation.
Let centroid coordinates be $(x,y)$, then \[ a=-\frac{3x}{\sqrt{3}},\quad b=-\frac{3y}{\sqrt{2}} \] Substitute into $3a^2+2b^2=16$, \[ 3\left(\frac{9x^2}{3}\right)+2\left(\frac{9y^2}{2}\right)=16 \] \[ 9x^2+9y^2=16 \Rightarrow x^2+y^2=\frac{16}{9} \] Step 4: Radius of the locus.
\[ \text{Radius}=\sqrt{\frac{16}{9}}=\frac{8}{3} \]