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Let a circle $\mathrm{C}$ of radius $\mathrm{5}$ lie below the x-axis. The line $\mathrm{L_{1}: 4 x+3 y+2=0}$ passes through the center $\mathrm{P}$ of the circle $\mathrm{C}$ and intersects the line $\mathrm{L_{2}: 3 x-4 y-11=0}$ at $\mathrm{Q}$ . The line $\mathrm{L_{2}}$ touches $\mathrm{C}$ at the point $\mathrm{Q}$ . Then the distance of $\mathrm{P}$ from the line $\mathrm{5 x-12 y+51=0}$ is __________.
Option: 1
11

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

Point of intersection of $\mathrm{L_{1}: 4 x+3 y+2=0}$ and $\mathrm{L_{2}: 3 x-4 y-11=0}$ is

$\mathrm{Q\left ( 1,-2 \right )}$ .

Centre $\mathrm{P}$ of the circle lies on $\mathrm{L_{1}}$

$\mathrm{Slope\: of\: L_{1}: \tan \theta=-\frac{4}{3} \Rightarrow \cos \theta: \frac{-3}{5}, \sin \theta=\frac{4}{5}}\\$

$\mathrm{Coordinates \: of \: P:(1-5 \cos \theta,-2 - 5 \sin \theta)}\\$

$\mathrm{=\left(4,-6\right)}$

$\mathrm{distance\: of \: P(4,-6)\: from \: 5 x-12 y+51\: is}$

$\mathrm{d=\left|\frac{5 \times 4-12 \times(-6)+51}{\sqrt{5^{2}+122}}\right|=\frac{20+72+51}{13}=\frac{143}{13}=11}$

Hence the answer is $\mathrm{11}$

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Rakesh

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