Let $P\left(x_1, y_1\right)$ be the point from which the tangents are drawn to the circles
$S_1 \equiv x^2+y^2-8 x+40=0$
$S_2 \equiv 5 x^2+5 y^2-25 x+80=0$
$S_3 \equiv x^2+y^2-8 x+16 y+160=0$
Since, the length of the tangent from $P$ to the circle $S_1, S_2, S_3$ are equal
$\therefore \quad \sqrt{S_1}=\sqrt{S_2}=\sqrt{S_3}$
$\Rightarrow \quad S_1=S_2=S_3$
$x_1^2+y_1^2-8 x_1+40=5 x_1^2+5 x_2^2-25 x_1+80$
$=x_1^2+y^2-8 x_1+16 y_1+160$ $\ldots$ (i)
Taking first and third part of above relation (i), we get
$-40+16 y_1+160=0$
$16 y_1+120=0$
$y_1=-\frac{120}{16} \Rightarrow-\frac{15}{2}$
Taking first and second part of the relation (i).
$-3 x_1+24=0$
$x_1=\frac{24}{3} \Rightarrow x_1=8$
Hence, the point $P$ is $\left(8, \frac{-15}{2}\right)$.