question_answer

From the point \[P(16,\,\,7)\] tangents \[PQ\] and \[PR\] are drawn to the circle\[{{x}^{2}}+{{y}^{2}}-2x-4y-20=0\].If\[c\]be the centre of the circle, then area of quadrilateral \[PQCR\] is

A) \[450\,\,sq\,\,unit\]                       

B) \[15\,\,sq\,\,unit\]

C) \[50\,\,sq\,\,unit\]                         

D) \[75\,\,sq\,\,unit\]

Correct Answer: D

Solution :

Key Idea If tangents are drawn from a point to the circle, then quadrilateral \[PQCR\] makes two equal right triangles. Given, equation of circle is                 \[{{x}^{2}}+{{y}^{2}}-2x-4y-20=0\] \[\therefore \]Centre is \[(1,\,\,2)\] and radius,                 \[r=\sqrt{{{1}^{2}}+{{2}^{2}}+20}\]                    \[=\sqrt{25}=5\] Now, \[PC=\sqrt{{{(16-1)}^{2}}+{{(7-2)}^{2}}}\]                    \[=\sqrt{225+25}\]                    \[=\sqrt{250}\] In\[\Delta PCQ\],            \[PQ=\sqrt{P{{C}^{2}}-Q{{C}^{2}}}\]                    \[=\sqrt{{{(250)}^{2}}-{{(5)}^{2}}}\]                    \[=\sqrt{250-25}\]                    \[=\sqrt{225}\]                    \[=15\] \[\therefore \]Area of quadrilateral\[PQCR\]                 \[=2\]area of\[\Delta PCQ\]                 \[=2\cdot \frac{1}{2}PQ\cdot QC\]                 \[=1\cdot 15\cdot 5\]                 \[=75\,\,sq\,\,unit\]