Question

$C_1$ and $C_2$ are two concentric circles, the radius of $C_2$ being twice that of $C_1$. From a point $P$ on $C_2$, tangents $PA$ and $PB$ are drawn to $C_1$. Prove that the centroid of the triangle $PAB$ lies on $C_1$.

Answer 1

Let $P(h,k)$ be on $C_2\therefore h^2+k^2=4^2......\left(1\right)$
C.C. of $P$ w.r.t. $C_1$ is $hx+ky=r^2$
It intersects $C_1,x^2+y^2=a^2$ is $A$ and $B$.
Eliminating $y$, we get
$x^2+{\left(\frac{r^2-hx}{k}\right)}^2=r^2$
or $x^2(h^2+k^2)-2r^{2}hx+r^4-r^2{k}^2=0$
or $x^2.4r^2-2r^{2}hx+r^2(r^2-k^2)=0$
$\therefore x_1+x_2=\frac{2r^{2}h}{4r^2}=\frac{h}{2},y_1+y_2=\frac{k}{2}$
If $(x,y)$ be the centroid of $△PAB$, then
$3x=x_1+x_2+h=\frac{h}{2}+h=\frac{3h}{2}$
$\therefore x=h/2$ or $h=2x$ and similarly $k=2y$
Putting in $\left(1\right)$ we get $4x^2+4y^2=4r^2$
$\therefore$ Locus is $x^2+y^2=r^2$ i.e. $C_1$.