Question

Let a variable point $P$ moves on the circle $x^2+y^2=9$ and the line $x+y=3$ cuts the co-ordinates axes at $A$ and $B$. Also locus of the centroid of $△PAB$ is the curve $C$. If the locus of the centre of a variable circle which touches the circle $x^2+y^2=9$ and curve $C$ is an ellipse whose eccentricity is e, find the value of $e^{-2}$.

Answer 1

Answer: 8

$h=\frac{3\cos \theta +0+3}{3}=\cos \theta +1$
$\therefore k=\frac{3\sin \theta +3+0}{3}=\sin \theta +1$
$\therefore (h-1{)}^2+(k-1{)}^2=1$
$(x-1{)}^2+(y-1{)}^2=1$
$x^2-2x+y^2-2y+1=0$
$C:x^2+y^2-2x-2y+1=0$

$C{C}_1=r+r_1$
$C{C}_2=r_2-r$
$C{C}_1+C{C}_2=r_1+r_2=1+3=4=2a$
Hence locus of $C$ is an ellipse.
Now $2a=4$ and $2ae=\sqrt{2}\Rightarrow e=\frac{\sqrt{2}}{4}=\frac{1}{2\sqrt{2}}$
Hence $\frac{1}{e^2}=8$