Question

The area of the region formed by x² + y² − 6x − 4y + 12 ≤ 0, y ≤ x and x ≤ 5/2 is ______ .

विकल्प

• \[\frac{\pi}{6} - \frac{\sqrt{3} + 1}{8}\]

• \[\frac{\pi}{6} + \frac{\sqrt{3} + 1}{8}\]

• \[\frac{\pi}{6} - \frac{\sqrt{3} - 1}{8}\]

• none of these

Answer 1

\[\frac{\pi}{6} - \frac{\sqrt{3} - 1}{8}\]

 

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,
\[A = \int_2^\frac{5}{2} \left( y_2 - y_1 \right) dx ............\left( \text{Where, }y_1 = - \sqrt{1 - \left( x - 3 \right)^2} + 2\text{ and }y_2 = x \right)\]
\[ = \int_2^\frac{5}{2} \left[ x - \left( - \sqrt{1 - \left( x - 3 \right)^2} + 2 \right) \right] d x\]
\[ = \int_2^\frac{5}{2} \left[ x + \sqrt{1 - \left( x - 3 \right)^2} - 2 \right] d x\]
\[ = \left[ \frac{x^2}{2} + \frac{\left( x - 3 \right)}{2}\sqrt{1 - \left( x - 3 \right)^2} + \frac{1}{2} \sin^{- 1} \left( x - 3 \right) - 2x \right]_2^\frac{5}{2} \]
\[ = \left[ \frac{\left( \frac{5}{2} \right)^2}{2} + \frac{\frac{5}{2} - 3}{2}\sqrt{1 - \left\{ \left( \frac{5}{2} \right) - 3 \right\}^2} + \frac{1}{2} \sin^{- 1} \left( \frac{5}{2} - 3 \right) - 2\left( \frac{5}{2} \right) \right] - \left[ \frac{2^2}{2} + \frac{2 - 3}{2}\sqrt{1 - \left( 2 - 3 \right)^2} + \frac{1}{2} \sin^{- 1} \left( 2 - 3 \right) - 2\left( 2 \right) \right]\]
\[ = \left[ \frac{25}{8} - \frac{1}{4}\sqrt{1 - \frac{1}{4}} + \frac{1}{2} \sin^{- 1} \left( - \frac{1}{2} \right) - 5 \right] - \left[ 2 - \frac{1}{2} \times 0 + \frac{1}{2} \sin^{- 1} \left( - 1 \right) - 4 \right]\]
\[ = \left[ - \frac{15}{8} - \frac{\sqrt{3}}{8} + \frac{1}{2} \times \left( - \frac{\pi}{6} \right) \right] - \left[ + \frac{1}{2} \times \left( - \frac{\pi}{2} \right) - 2 \right]\]
\[ = - \frac{15}{8} - \frac{\sqrt{3}}{8} - \frac{\pi}{12} + \frac{\pi}{4} + 2\]
\[ = \frac{\pi}{6} - \frac{\sqrt{3} - 1}{8}\]