The lengths of the tangent drawn from any point on the circle $15 x^{2}+15 y^{2}-48 x+64 y=0$ to the two…
The lengths of the tangent drawn from any point
on the circle $15 x^{2}+15 y^{2}-48 x+64 y=0$ to the two circles $5 x^{2}+5 y^{2}-24 x+32 y+75=0$ and $5 x^{2}+5 y^{2}-48 x$ $+64 y+300=0$ are in the ratio of
1: 2
2: 3
3: 4
None
Solution
Let \(\mathrm{P}(\mathrm{h}, \mathrm{k})\) be a point on the circle
\(15 x^{2}+15 y^{2}-48 x+64 y=0\)
Then the lengths of the tangents from \(\mathrm{P}(\mathrm{h}, \mathrm{k})\) to
\(\begin{aligned} &5 x^{2}+5 y^{2}-24 x+32 y+75=0 \\ &5 x^{2}+5 y^{2}-48 x+64 y+300=0 \text { are } \\ &P T_{1}=\sqrt{h^{2}+k^{2}-\frac{24}{5} h+\frac{32}{5} k+15} \\ &\text { and } P T_{2}=\sqrt{h^{2}+k^{2}-\frac{48}{5} h+\frac{64}{5} k+60} \\ &\text { or } P T_{1}=\sqrt{\frac{48}{15} h-\frac{64}{15} k-\frac{24}{5} h+\frac{32}{5} k+15}=\sqrt{\frac{32}{15} k-\frac{24}{15} h+15} \end{aligned}\)
(Since \((\mathrm{h}, \mathrm{k})\) lies on \(15 \mathrm{x}^{2}-15 \mathrm{y}^{2}-48 \mathrm{x}+64 \mathrm{y}=0\)
\(\begin{aligned} &\left.\therefore \mathrm{h}^{2}+\mathrm{k}^{2}-\frac{48}{15} \mathrm{~h}+\frac{64}{15} \mathrm{k}=0\right) \\ &\text { and } \mathrm{PT}_{2}=\sqrt{\frac{48}{15} \mathrm{~h}-\frac{64}{5} \mathrm{k}-\frac{48}{5} \mathrm{~h}+\frac{64}{5} \mathrm{k}+60} \\ &=\sqrt{-\frac{96}{15} \mathrm{~h}+\frac{128}{15} \mathrm{k}+60}=2 \sqrt{-\frac{24}{15} \mathrm{~h}+\frac{32}{15} \mathrm{k}+15}=2 \mathrm{PT}_{1} \\ &\Rightarrow \mathrm{PT}_{1}: \mathrm{PT}_{2}=1: 2 \end{aligned}\)