Question
Let $$AB$$ be a chord of the circle $${x^2} + {y^2} = {r^2}$$ subtending a right angle at the centre. Then the locus of the centroid of the triangle $$PAB$$ as $$P$$ moves on the circle is-
A.
a parabola
B.
a circle
C.
an ellipse
D.
a pair of straight lines
Answer :
a circle
Solution :
$${x^2} + {y^2} = {r^2}$$ is a circle with centre at $$\left( {0,\,0} \right)$$ and radius $$r$$ units.
Any arbitrary point $$P$$ on it is $$\left( {r\,\cos \,\theta ,\,r\,\sin \,\theta } \right)$$
Choosing $$A$$ and $$B$$ as $$\left( { - r,\,0} \right)$$ and $$\left( { 0,\,- r} \right)$$
[So that $$\angle AOB = {90^ \circ }$$ ]
For locus of centroid of $$\Delta ABP$$
$$\eqalign{ & \left( {\frac{{r\,\cos \,\theta - r}}{3},\,\frac{{r\,\sin \,\theta - r}}{3}} \right) = \left( {x,\,y} \right) \cr & \Rightarrow r\,\cos \,\theta - r = 3x \cr & \,\,\,\,\,\,r\,\sin \,\theta - r = 3y \cr & \Rightarrow r\,\cos \,\theta = 3x + r \cr & \,\,\,\,\,\,r\,\sin \,\theta = 3y + r \cr} $$
Squaring and adding $${\left( {3x + r} \right)^2} + {\left( {3y + r} \right)^2} = {r^2}$$ which is a circle.
$${x^2} + {y^2} = {r^2}$$ is a circle with centre at $$\left( {0,\,0} \right)$$ and radius $$r$$ units.
Any arbitrary point $$P$$ on it is $$\left( {r\,\cos \,\theta ,\,r\,\sin \,\theta } \right)$$
Choosing $$A$$ and $$B$$ as $$\left( { - r,\,0} \right)$$ and $$\left( { 0,\,- r} \right)$$
[So that $$\angle AOB = {90^ \circ }$$ ]
For locus of centroid of $$\Delta ABP$$
$$\eqalign{ & \left( {\frac{{r\,\cos \,\theta - r}}{3},\,\frac{{r\,\sin \,\theta - r}}{3}} \right) = \left( {x,\,y} \right) \cr & \Rightarrow r\,\cos \,\theta - r = 3x \cr & \,\,\,\,\,\,r\,\sin \,\theta - r = 3y \cr & \Rightarrow r\,\cos \,\theta = 3x + r \cr & \,\,\,\,\,\,r\,\sin \,\theta = 3y + r \cr} $$
Squaring and adding $${\left( {3x + r} \right)^2} + {\left( {3y + r} \right)^2} = {r^2}$$ which is a circle.