Question

The length of the tangents 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

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

Answer 1

Answer: (A)

Let any point $P\left(x_1,y_1\right)$ to the circle $x^2+y^2-\frac{16x}{5}+\frac{64y}{15}=0$
$\Rightarrow x_1{}^2+y_1{}^2-\frac{16x_1}{5}+\frac{64y_1}{15}=0$
Length of tangent from $P\left(x_1,y_1\right)$ to the circle are in ratio
$\frac{\sqrt{S_1}}{\sqrt{S_2}}=\frac{\sqrt{x_1^2+y_1^2-\frac{24}{5}{x}_1+\frac{32}{5}{y}_1+15}}{\sqrt{x_1^2+y_1^2-\frac{48}{5}{x}_1+\frac{64}{5}{y}_1+60}}$
$=\sqrt{\frac{\frac{16}{5}{x}_1-\frac{64}{15}{y}_1-\frac{24}{5}{x}_1+\frac{32}{5}{y}_1+15}{\frac{16}{5}{x}_1-\frac{64}{15}{y}_1-\frac{48}{5}{x}_1+\frac{64}{5}{y}_1+60}}$
$=\sqrt{\frac{-24x_1+32y_1+225}{-96x_1+128y_1+900}}$
$=\sqrt{\frac{-24x_1+32y_1+225}{4\left(-24x_1+32y_1+225\right)}}=\frac{1}{2}$