The point which has the same power with respect to each of the circles x 2 + y 2 - 8 x + 40 = 0 ,   x 2…

Given circle equations 

$x^2+y^2-8x+40=0---\left(I\right)$

$x^2+y^2-5x+16=0----\left(II\right)$

$x^2+y^2-8x+16y+160=0----\left(III\right)$ 

Power of point with respect to circle $=S_1$

For point $P\left(8, \frac{-15}{2}\right)$

With respect to equation $\left(I\right)$

$=64+\frac{225}{4}-64+40=\frac{385}{4}$

With respect to equation $\left(II\right)$

$= 64+\frac{225}{4}-40+16=\frac{385}{4}$

With respect to third circle

$=\frac{385}{4}$

The remaining points in other options are not having same power.

So required point is $\left(8, \frac{-15}{2}\right)$.