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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}

Posted by

Rakesh

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