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\[\Delta PCQ\] \[=2\cdot \frac{1}{2}PQ\cdot QC\] \[=1\cdot 15\cdot 5\] \[=75\,\,sq\,\,unit\]You need to login to perform this action.
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