Question

The area of the region formed by x² + y² − 6x − 4y + 12 ≤ 0, y ≤ x and x ≤ 5/2 is
(a) $\frac{\mathrm{\pi }}{6}-\frac{\sqrt{3}+1}{8}$

(b) $\frac{\mathrm{\pi }}{6}+\frac{\sqrt{3}+1}{8}$

(c) $\frac{\mathrm{\pi }}{6}-\frac{\sqrt{3}-1}{8}$

(d) none of these

Answer 1

$\left(\mathrm{c}\right)\frac{\mathrm{\pi }}{6}-\frac{\sqrt{3}-1}{8}$



$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ {x}^2+y^2-6x-4y+12\le 0 \\ y\le x \\ x\le \frac{5}{2} \\ \mathrm{Following}\mathrm{are}\mathrm{the}\mathrm{corresponding}\mathrm{equations}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{inequation}. \\ {x}^2+y^2-6x-4y+12=0.....\left(1\right) \\ y=x.....\left(2\right) \\ x=\frac{5}{2}.....\left(3\right) \end{gathered}$$
Here, ABC is our required region in which point A is intersection of (1) and (3), point B is intersection of (1) and (2) and point C is intersection of (2) and (3).
By solving (1), (2) and (3) we get the coordinates of B and C as
$$\begin{gathered} B\equiv \left(2,2\right) \\ C\equiv \left(\frac{5}{2},\frac{5}{2}\right) \end{gathered}$$
Now, the equation of the circle is,
$$\begin{gathered} x^2+y^2-6x-4y+12=0 \\ \Rightarrow {\left(x-3\right)}^2+{\left(y-2\right)}^2=1 \\ \Rightarrow {\left(y-2\right)}^2=1-{\left(x-3\right)}^2 \\ \Rightarrow y-2=\pm \sqrt{1-{\left(x-3\right)}^2} \\ \Rightarrow y=\pm \sqrt{1-{\left(x-3\right)}^2}+2 \\ \Rightarrow y=\sqrt{1-{\left(x-3\right)}^2}+2\mathrm{or}-\sqrt{1-{\left(x-3\right)}^2}+2 \\ y=\sqrt{1-{\left(x-3\right)}^2}+2\mathrm{is}\mathrm{not}\mathrm{possible}, \\ \mathrm{Therefore},\mathrm{y}=-\sqrt{1-{\left(\mathrm{x}-3\right)}^2}+2 \end{gathered}$$
The area of the required region ABC,
$$\begin{gathered} A={\int }_2^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(y_2-y_1\right)\mathit{d}x\left(\mathrm{Where},y_1=-\sqrt{1-{\left(x-3\right)}^2}+2\mathrm{and}{y}_2=x\right) \\ ={\int }_2^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[x-\left(-\sqrt{1-{\left(x-3\right)}^2}+2\right)\right]\mathit{d}x \\ ={\int }_2^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[x+\sqrt{1-{\left(x-3\right)}^2}-2\right]\mathit{d}x \\ ={\left[\frac{x^2}{2}+\frac{\left(x-3\right)}{2}\sqrt{1-{\left(x-3\right)}^2}+\frac{1}{2}{\sin }^{-1}\left(x-3\right)-2x\right]}_2^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =\left[\frac{{\left({\displaystyle \frac{5}{2}}\right)}^2}{2}+\frac{{\displaystyle \frac{5}{2}-3}}{2}\sqrt{1-{\left\{\left(\frac{5}{2}\right)-3\right\}}^2}+\frac{1}{2}{\sin }^{-1}\left(\frac{5}{2}-3\right)-2\left(\frac{5}{2}\right)\right]-\left[\frac{2^2}{2}+\frac{2-3}{2}\sqrt{1-{\left(2-3\right)}^2}+\frac{1}{2}{\sin }^{-1}\left(2-3\right)-2\left(2\right)\right] \\ =\left[\frac{25}{8}-\frac{1}{4}\sqrt{1-\frac{1}{4}}+\frac{1}{2}{\sin }^{-1}\left(-\frac{1}{2}\right)-5\right]-\left[2-\frac{1}{2}\times 0+\frac{1}{2}{\sin }^{-1}\left(-1\right)-4\right] \\ =\left[-\frac{15}{8}-\frac{\sqrt{3}}{8}+\frac{1}{2}\times \left(-\frac{\mathrm{\pi }}{6}\right)\right]-\left[+\frac{1}{2}\times \left(-\frac{\mathrm{\pi }}{2}\right)-2\right] \\ =-\frac{15}{8}-\frac{\sqrt{3}}{8}-\frac{\mathrm{\pi }}{12}+\frac{\mathrm{\pi }}{4}+2 \\ =\frac{\mathrm{\pi }}{6}-\frac{\sqrt{3}-1}{8} \end{gathered}$$