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. $Noneofthese$

Answer 1

The correct option is A $1:2$

Let $P(h,k)$ be a point on the circle
$15x^2+15y^2-48x+64y=0$
Then the lengths of the tangents from $P(h,k)$ to
$5x^2+5y^2-24x+32y+75=0$
$5x^2+5y^2-48x+64y+300=0$ are
$P{T}_1=\sqrt{h^2+k^2-\frac{24}{5}h+\frac{32}{5}k+15}$
and $P{T}_2=\sqrt{h^2+k^2-\frac{48}{5}h+\frac{64}{5}k+60}$
or $P{T}_1=\sqrt{\frac{48}{15}h-\frac{64}{15}k-\frac{24}{5}h+\frac{32}{5}k+15}=\sqrt{\frac{32}{15}k-\frac{24}{15}h+15}$
(Since $(h,k)$ lies on $15x^2-15y^2-48x+64y=0$
$\therefore h^2+k^2-\frac{48}{15}h+\frac{64}{15}k=0)$
and $P{T}_2=\sqrt{\frac{48}{15}h-\frac{64}{5}k-\frac{48}{5}h+\frac{64}{5}k+60}$
$=\sqrt{-\frac{96}{15}h+\frac{128}{15}k+60}=2\sqrt{-\frac{24}{15}h+\frac{32}{15}k+15}=2P{T}_1$
$\Rightarrow P{T}_1:P{T}_2=1:2$