Question

Let $AB$ be a chord of the circle $x^2+y^2=r^2$ subtending right angle at the center. Then the locus of the centroid of the triangle $PAB$ as $P$ moves on the circle is

• A. a circle
• B. a parabola
• C. an ellipse
• D. a pair of straight lines

Answer 1

The correct option is A a circle

Equation of the circle is $x^2+y^2=r^2.$

Let co-ordinates of the point $P(x_1,y_1)=(r\cos \theta ,r\sin \theta ).$

Since chord $AB$ subtends right angle at the centre, therefore coordinates of point $A(x_2,y_2)=(r,0)$ and point $B(x_3,y_3)=(0,r).$

We know that coordinates of the centroid of the $△PAB$

$(h,k)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})=(\frac{r\cos \theta +r+0}{3},\frac{r\sin \theta +0+r}{3})$

$=(\frac{1}{3}(r+r\cos \theta ),\frac{1}{3}(r+r\sin \theta ))$

Therefore, $h=\frac{1}{3}(r+r\cos \theta )\Rightarrow h-\frac{1}{3}r=\frac{1}{3}r\cos \theta$ ...(i)

and $k=\frac{1}{3}(r+r\sin \theta )\Rightarrow k-\frac{1}{3}r=\frac{1}{3}r\sin \theta$ ...(ii)

Squaring and adding Eqs. (i) and (ii), we get

${(h-\frac{1}{3}r)}^2+{(k-\frac{1}{3}r)}^2=\frac{1}{9}{r}^2({\cos }^2\theta +{\sin }^2\theta )=\frac{1}{9}{r}^2$

Thus, the locus of the centroid of $△PAB$ is ${(x-\frac{r}{3})}^2+{(y-\frac{r}{3})}^2=\frac{r^2}{9},$ which is a circle.