Question

If a circle of constant radius $3k$ passes through the origin $O$ and meets co-ordinate axes at $A$ and $B$, then the locus of the centroid of the triangle $OAB$ is

• A. $x^2+y^2=(2k{)}^2$
• B. $x^2+y^2=(3k{)}^2$
• C. $x^2+y^2=(4k{)}^2$
• D. $x^2+y^2=(6k{)}^2$

Answer 1

The correct option is A $x^2+y^2=(2k{)}^2$


Let coordinates of $A$ be $(x_1,0)$ and $B$ be $(0,y_1)$ and centroid of triangle $OAB$ be $(p,q)$
So,
$p=\frac{x_1+0+0}{3}\Rightarrow x_1=3pq=\frac{y_1+0+0}{3}\Rightarrow y_1=3q$

Hence, $A=(3p,0)$ and $B=(0,3q)$
As $OA$ and $OB$ are perpendicular, so $AB$ is the diameter of the circle,
$AB=6k\Rightarrow 9p^2+9q^2=36k^2\Rightarrow p^2+q^2=4k^2$

Hence, the locus of the centre is
$x^2+y^2=(2k{)}^2$