The lengths of the tangents from any point on  the circle $15x^2+15y^2-48x+64y=0$ to the circles

$5x^2+5y^2-24x+32y+75=0$

$5x^2+5y^2-48x+64y+300=0$ are in the ratio

Let $P (h, k)$be a point on the circle

$15x^2+15y^2-48x+64y=0$

Then $15h^2+15k^2-48h+64k=0$

$\therefore h^2+k^2=\frac{48}{15}h-\frac{64}{15}k$             …(i)

Let $P{T}_1$ and $P{T}_2$ be the lengths of the tangents from $P (h, k)$ to 

$5x^2+5y^2-24x+32y+75=0$

and, $5x^2+5y^2-48x+64y+300=0$ respectively. Then,

$P{T}_1=\sqrt{h^2+k^2-\frac{24}{5}h+\frac{32}{5}k+15}$

and, $P{T}_2=\sqrt{h^2+k^2-\frac{48}{5}h+\frac{64}{5}k+60}$

$\Rightarrow P{T}_1=\sqrt{\frac{48}{15}h}-\frac{64}{15}k-\frac{24}{5}h+\frac{32}{5}k+15 \text{ [Using (i)] }$

$\Rightarrow P{T}_1=\sqrt{\frac{32k}{15}-\frac{24}{15}h+15}$

and, $P{T}_2=\sqrt{\frac{48}{15}h-\frac{64}{15}k-\frac{48}{5}h+\frac{64}{5}k+60} \text{ [Using (i)] }$

$\begin{array}{l}\Rightarrow P{T}_2=\sqrt{-\frac{96}{15}h+\frac{128}{15}k+60}\\ \Rightarrow P{T}_2=2\sqrt{-\frac{24}{15}h+\frac{32}{15}k+15}=2P{T}_1\\ \therefore P{T}_1:P{T}_2=1:2\end{array}$