The area of the region formed by x2 + y2 − 6x − 4y + 12 $\leq$ 0, y $\leq$ x and x $\leq$ $\frac{5}{2}$ is ______ .

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$\pi$/6 - $\sqrt$(3) - $\frac{1}{8}$ We have, x^2 + y^2 - 6x - 4y + 12 \leq 0 y \leq x x \leq $\frac{5}{2}$ Following are the corresponding equations of the given inequation . x^2 + y^2 - 6x - 4y + 12 = 0 . . . . . \left( 1 \right) y = x . . . . . \left( 2 \right) x = $\frac{5}{2}$ . . . . . \left( 3 \right) 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 B \equiv \left( 2, 2 \right) C \equiv \left( $\frac{5}{2}$, $\frac{5}{2}$ \right) Now, the equation of the circle is, 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 or - \sqrt{1 - \left( x - 3 \right)^2} + 2 y = \sqrt{1 - \left( x - 3 \right)^2} + 2\text{ is not possible,} \text{ Therefore, }y = - \sqrt{1 - \left( x - 3 \right)^2} + 2 The area of the required region ABC, A = \int_2^$\frac{5}{2}$ \left( y_2 - y_1 \right) dx ............\left( \text{Where, }y_1 = - \sqrt{1 - \left( x - 3 \right)^2} + 2\text{ and }y_2 = x \right) = \int_2^$\frac{5}{2}$ \left[ x - \left( - \sqrt{1 - \left( x - 3 \right)^2} + 2 \right) \right] d x = \int_2^$\frac{5}{2}$ \left[ x + \sqrt{1 - \left( x - 3 \right)^2} - 2 \right] d x = \left[ x^$\frac{2}{2}$ + \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^$\frac{5}{2}$ = \left[ \frac{\left( $\frac{5}{2}$ \right)^2}{2} + $\frac{5}{2}$ - $\frac{3}{2}$\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[ 2^$\frac{2}{2}$ + 2 - $\frac{3}{2}$\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[ - $\frac{15}{8}$ - $\sqrt$(3)/8 + $\frac{1}{2}$ $\times$ \left( - $\pi$/6 \right) \right] - \left[ + $\frac{1}{2}$ $\times$ \left( - $\pi$/2 \right) - 2 \right] = - $\frac{15}{8}$ - $\sqrt$(3)/8 - $\pi$/12 + $\pi$/4 + 2 = $\pi$/6 - $\sqrt$(3) - $\frac{1}{8}$ [Solution diagram: 43956_solution_1.png]

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