JEE MAIN - Mathematics (2022 - 29th July Evening Shift - No. 19)
Let $$A B$$ be a chord of length 12 of the circle $$(x-2)^{2}+(y+1)^{2}=\frac{169}{4}$$. If tangents drawn to the circle at points $$A$$ and $$B$$ intersect at the point $$P$$, then five times the distance of point $$P$$ from chord $$A B$$ is equal to __________.
Answer
72
Explanation
Here $A M=B M=6$
$$ O M=\sqrt{\left(\frac{13}{2}\right)^{2}-6^{2}}=\frac{5}{2} $$
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$$ \sin \theta=\frac{12}{13} $$
In $\triangle P A O$ :
$$ \begin{aligned} &\frac{P O}{O A}=\sec \theta \\\\ &P O=\frac{13}{2} \cdot \frac{13}{5}=\frac{169}{10} \\\\ &\therefore P M=\frac{169}{10}-\frac{5}{2}=\frac{144}{10}=\frac{72}{5} \\\\ &\therefore 5 P M=72 . \end{aligned} $$
$$ O M=\sqrt{\left(\frac{13}{2}\right)^{2}-6^{2}}=\frac{5}{2} $$
$$ \sin \theta=\frac{12}{13} $$
In $\triangle P A O$ :
$$ \begin{aligned} &\frac{P O}{O A}=\sec \theta \\\\ &P O=\frac{13}{2} \cdot \frac{13}{5}=\frac{169}{10} \\\\ &\therefore P M=\frac{169}{10}-\frac{5}{2}=\frac{144}{10}=\frac{72}{5} \\\\ &\therefore 5 P M=72 . \end{aligned} $$
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