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Question

The area of the region formed by x2 + y2 − 6x − 4y + 12 ≤ 0, y ≤ x and x ≤ 5/2 is
(a) π6-3+18

(b) π6+3+18

(c) π6-3-18

(d) none of these

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Solution

c π6-3-18



We have,x2+y2-6x-4y+120yxx52Following are the corresponding equations of the given inequation.x2+y2-6x-4y+12=0 .....1y=x .....2x=52 .....3
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
B2, 2C52, 52
Now, the equation of the circle is,
x2+y2-6x-4y+12=0x-32+y-22=1y-22=1-x-32y-2=±1-x-32y=±1-x-32+2y=1-x-32 +2 or -1-x-32+ 2y=1-x-32 +2 is not possible,Therefore, y=-1-x-32+ 2
The area of the required region ABC,
A=252y2-y1 dx Where, y1=-1-x-32+2 and y2=x=252x--1-x-32+2 dx=252x+1-x-32-2 dx=x22+x-321-x-32+12sin-1x-3-2x252=5222+52-321-52-32+12sin-152-3-252-222+2-321-2-32+12sin-12-3-22=258-141-14+12sin-1-12-5-2-12×0+12sin-1-1-4=-158-38+12×-π6-+12×-π2-2=-158-38-π12+π4+2=π6-3-18

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