`C_(1) and C_(2)` are two concentrate 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 `DeltaPAB` lies on `C_(1)`

Explanation

Let the coordinates of point P be `(2r cos theta, 2r sin theta)`
We have , OA = r, OP = 2r
Since, `Delta OAP` is a right angled triangle.

`cosphi=1//2rArr phi=pi//3`
`therefore` Coordinates of A are `{cos(theta-pi//3),rsin(theta-pi//3)}`
If p, q is the centroid of Delta PAB, then
`p=(1)/(3)[rcos(theta-pi//3)+rcos(theta+pi//3)+2rcostheta]`
`=(1)/(3)[r(2"cos"(theta-(pi)/(3)+theta+(pi)/(3))/(2)*"cos"(theta-(pi)/(3)-theta-(pi)/(3))/(2))+2rcostheta]`
`(1)/(3)[r{2costhetapi//3}+2rcostheta]`
`=(1)/(3)[r*costheta+2rcostheta]=rcostheta`
and `q=(1)/(3)[rsin( theta-(pi)/(3))+rsin(theta+(pi)/(3))+2rsintheta]`
`=(1)/(3)[r(2"sin"(theta-(pi)/(3)+theta+(pi)/(3))/(2)*"cos"(theta-(pi)/(3)-theta-(pi)/(3))/(2))+2rsintheta]`
`=(1)/(3)[rsin(2sinthetacospi//3)+2rsintheta]`
`=(1)/(3)[r(sintheta)+2rsintheta]=rsintheta`
Now, (p, q) = `(r cos theta, r sin theta) " lies on " x^(2) + y^(2) = r^(2)` which is called `C_(1)`.