Question

Given triangle $OAB$ where $O$ is the origin, $A=(0,-\sqrt{3}a)$ and $B=(-\sqrt{2}b,0)$. Let the circumradius of $\triangle OAB$ be $4$ units. If the locus of the centroid of $\triangle OAB$ is a circle, then its radius is:

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

Hint:

The locus of the centroid scales directly with the coordinates of the vertices, shrinking distances by a factor of $\frac{1}{3}$.

Answer 1

The Correct Option is A

Step 1: Write coordinates of the centroid.
The centroid $G$ of triangle with vertices $O(0,0)$, $A(0,-\sqrt{3}a)$, and $B(-\sqrt{2}b,0)$ is: \[ G\left(\frac{-\sqrt{2}b}{3},\frac{-\sqrt{3}a}{3}\right) \] Step 2: Use the formula for circumradius.
The circumradius $R$ of a triangle with vertices at $(0,0)$, $(x_1,y_1)$, $(x_2,y_2)$ is: \[ R = \frac{\sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2}\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}{4\Delta} \] Given $R=4$, simplifying using the given coordinates leads to: \[ a^2 + b^2 = 16 \] Step 3: Write the locus of the centroid.
\[ x = -\frac{\sqrt{2}b}{3}, \quad y = -\frac{\sqrt{3}a}{3} \] \[ \Rightarrow \frac{9x^2}{2} + \frac{9y^2}{3} = a^2 + b^2 \] Step 4: Substitute the given condition.
\[ \frac{9x^2}{2} + 3y^2 = 16 \] \[ x^2 + y^2 = \frac{64}{9} \] Step 5: Identify the radius of the locus.
\[ \text{Radius} = \sqrt{\frac{64}{9}} = \frac{8}{3} \]