Q. 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
Jharkhand CECEJharkhand CECE 2008
Solution:
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 $
