First slide

Definition of a circle

Question

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

Moderate

A

$1 : 2$
B

$2 : 3$
C

$3 : 4$
D

none of these

Solution

Let $P (h, k)$ be a point on the circle
$15 x^{2} + 15 y^{2} - 48 x + 64 y = 0$
Then $15 h^{2} + 15 k^{2} - 48 h + 64 k = 0$
$∴ h^{2} + k^{2} = \frac{48}{15} h - \frac{64}{15} k$ …(i)
Let $P T_{1}$ and $P T_{2}$ be the lengths of the tangents from $P (h, k)$ to
$5 x^{2} + 5 y^{2} - 24 x + 32 y + 75 = 0$
and, $5 x^{2} + 5 y^{2} - 48 x + 64 y + 300 = 0$ respectively. Then,
$P T_{1} = \sqrt{h^{2} + k^{2} - \frac{24}{5} h + \frac{32}{5} k + 15}$
and, $P T_{2} = \sqrt{h^{2} + k^{2} - \frac{48}{5} h + \frac{64}{5} k + 60}$
$\Rightarrow P T_{1} = \sqrt{\frac{48}{15} h} - \frac{64}{15} k - \frac{24}{5} h + \frac{32}{5} k + 15 [Using (i)]$
$\Rightarrow P T_{1} = \sqrt{\frac{32 k}{15} - \frac{24}{15} h + 15}$
and, $P T_{2} = \sqrt{\frac{48}{15} h - \frac{64}{15} k - \frac{48}{5} h + \frac{64}{5} k + 60} [Using (i)]$
$\begin{array}{l} \Rightarrow P T_{2} = \sqrt{- \frac{96}{15} h + \frac{128}{15} k + 60} \\ \Rightarrow P T_{2} = 2 \sqrt{- \frac{24}{15} h + \frac{32}{15} k + 15} = 2 P T_{1} \\ ∴ P T_{1} : P T_{2} = 1 : 2 \end{array}$

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