Question

The area of the region $\text{(in sq.units)}$ formed by $x^2+y^2-6x-4y+12\le 0,y\le x$ and $x\le \frac{5}{2}$ is

• A. $(\frac{1}{8}+\frac{\sqrt{3}}{8}-\frac{\pi }{6})$
• B. $(\frac{1}{8}-\frac{\sqrt{3}}{8}+\frac{\pi }{6})$
• C. $(\frac{1}{8}-\frac{\sqrt{3}}{8}-\frac{\pi }{6})$
• D. None of these

Answer 1

The correct option is B $(\frac{1}{8}-\frac{\sqrt{3}}{8}+\frac{\pi }{6})$

Given curve is

$x^2+y^2-6x-4y+12\le 0,y\le x,x\le 5/2\Rightarrow (x-3{)}^2+(y-2{)}^2\le 1,y\le x,x\le 5/2\Rightarrow (y-2{)}^2=\{1-(x-3{)}^2\}\Rightarrow y=2\pm \sqrt{1-(x-3{)}^2}\Rightarrow y=2-\sqrt{1-(x-3{)}^2}\text{as}x\le \frac{5}{2}$

Hence, the required area

$=\frac{1}{2}(2+\frac{5}{2})\frac{1}{2}-{\int }_2^{5/2}(2-\sqrt{1-(x-3{)}^2})dx=\frac{9}{8}-{\int }_2^{5/2}(2-\sqrt{1-(x-3{)}^2})dx=\frac{9}{8}-{(2x-\frac{(x-3)}{2}\sqrt{1-(x-3{)}^2}-\frac{1}{2}{\sin }^{-1}(x-3\left)\right)}_2^{5/2}=\frac{\pi }{6}+\frac{1-\sqrt{3}}{8}$