Question

The lengths of the tangent drawn from any point on the circle $15x^2 + 15x^2 - 48x +64y = 0$ 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. 1:02
• B. 2:03
• C. 3:04
• D. None of these

Answer 1

Answer: (A)

1:02

Answer 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 $5x^2 + 5y^2 -24x + 32y +75 = 0$ and $5x^2 + 5y^2 - 48x + 64y +300 = 0$ are $PT_{1} = \sqrt{h^{2}+k^{2} - \frac{24}{5}h + \frac{32}{5}k +15}$ and $PT_{2} = \sqrt{h^{2}+k^{2} - \frac{48}{5}h + \frac{64}{5}k +30}$ 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 $\left(h, k\right)$ lies on $5x^{2} + 5y^{2} - 48x + 64y = 0$ $\therefore h^{2}+k^{2} - \frac{48}{15}h + \frac{64}{15}k = 0$ and $PT_{2} = \sqrt{\frac{48}{15}h - \frac{64}{15}k-\frac{48}{5}h - \frac{64}{5}k +60}$ $= \sqrt{-\frac{96}{15}h+\frac{128}{15}k +60 } = 2\sqrt{-\frac{24}{15}h}+\frac{32}{15}k + 15 = 2PT_{1}$ $\Rightarrow PT_{1} : PT_{2} =1 : 2$