Question

The lengths of the tangent drawn from any point on the circle $15x^2 +15y^2 - 48x + 64y = 0$ to the two circles $5x^2 + 5y^2 - 24x + 32y + 75 = 0$ and $5x^2 + 5y^2 - 48x + 64y + 300 = 0$ are in the ratio of

• A. 1 : 2
• B. 2 : 3
• C. 3 : 4
• D. None

Answer 1

Answer: (A)

Let $P ( h , k )$ be a point on the circle
$15 x^{2}+15 y^{2}-48 x+64 y=0$
Then the lengths of the tangents from $P ( h , k )$ to
$5 x ^{2}+5 y ^{2}-24 x +32 y +75=0$
$5 x ^{2}+5 y ^{2}-48 x +64 y +300=0 $ are
$PT _{1}=\sqrt{ h ^{2}+ k ^{2}-\frac{24}{5} h +\frac{32}{5} k +15}$
and $ PT _{2}=\sqrt{ h ^{2}+ k ^{2}-\frac{48}{5} h +\frac{64}{5} k +60}$
or $PT _{1}=\sqrt{\frac{48}{15} h -\frac{64}{15} k -\frac{24}{5} h +\frac{32}{5} k +15}=\sqrt{\frac{32}{15} k -\frac{24}{15} h +15}$
(Since $(h, k)$ lies on $15x^2-15 y ^{2}-48 x +64 y =0$
$\left.\therefore h ^{2}+ k ^{2}-\frac{48}{15} h +\frac{64}{15} k =0\right)$
and $PT _{2}=\sqrt{\frac{48}{15} h -\frac{64}{5} k -\frac{48}{5} h +\frac{64}{5} k +60}$
$=\sqrt{-\frac{96}{15} h +\frac{128}{15} k +60}=2 \sqrt{-\frac{24}{15} h +\frac{32}{15} k +15}=2 PT _{1}$
$\Rightarrow PT _{1}: PT _{2}=1: 2$