Question

Prove that the locus of the centroid of the triangle whose vertices are $\left(a\cos t,a\sin t,\right)$$\left(b\sin t,-b\cos \right),$and $\left(1,0\right)$,where t is a parameter, is circle.

Answer 1

Let locus of centroid be at $(h,k)$

then

$h=\frac{1+a\cos t+b\sin t}{3}$

$k=\frac{a\sin t-b\cos t}{3}$

$3h-1=a\cos t+b\sin t$ ______ (1)

$3k=a\sin t-b\cos t$ __________ (2)

square both equations and add

$(3h-1{)}^2+(3k{)}^2=a^2({\cos }^{2}t+{\sin }^{2}t)+b^2({\sin }^{2}t+{\cos }^{2}t)$

$(3h-1{)}^2+(3k{)}^2=a^2+b^2$

put $h=x,k=y$

$(3x-1{)}^2+9y^2=a^2+b^2$

it is a circle

with center $(\frac{1}{3},0)$ and radius $\sqrt{\frac{a^2+b^2}{9}}=\sqrt{\frac{a^2+b^2}{3}}$.