question_answer

The area of the region formed by\[{{x}^{2}}+{{y}^{2}}-6x-4y+12\le 0,y\le x\] and \[x\le \frac{5}{2}\] is

A) \[\left( \frac{\pi }{6}-\frac{\sqrt{3}+1}{8} \right)sq\,\,unit\]

B) \[\left( \frac{\pi }{6}+\frac{\sqrt{3}-1}{8} \right)sq\,\,unit\]

C) \[\left( \frac{\pi }{6}-\frac{\sqrt{3}-1}{8} \right)sq\,\,unit\]

D) None of theses

Correct Answer: C

Solution :

[c] The required area \[=\int_{2}^{5/2}{xdx-\int_{2}^{5/2}{\left[ 2-\sqrt{1-{{(x-3)}^{2}}} \right]dx}}\] \[=\left[ \frac{{{x}^{2}}}{2} \right]_{2}^{5/2}-[2x]_{2}^{5/2}+\] \[\left[ \frac{x-3}{2}\sqrt{1-{{(x-3)}^{2}}}+\frac{1}{2}{{\sin }^{-1}}(x-3) \right]_{2}^{5/2}\]  \[=\frac{9}{8}-1+\left( -\frac{\sqrt{3}}{8}+\frac{\pi }{6} \right)=\frac{\pi }{6}-\frac{\sqrt{3}-1}{8}\] sq. unit.