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in Definite Integrals by (29.1k points)
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The area of the region formed by x2 + y2 – 6x – 4y + 12 ≤ 0, y ≤ x and x ≤ 5/2 is

A. \(\frac{\pi}{6}\)\(-\frac{\sqrt{3}+1}{8}\)

B.  \(\frac{\pi}{6}\)\(+\frac{\sqrt{3}+1}{8}\)

C.  \(\frac{\pi}{6}\)\(-\frac{\sqrt{3}-1}{8}\)

D. None of these

1 Answer

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by (28.7k points)
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Best answer

Correct answer is D.

x2 + y2 – 6x – 4y + 12 ≤ 0 can be written as –

x2 – 6x + 9 – 9 + y2 – 4y + 4 – 4 + 12 ≤ 0

i.e., (x - 3)2 + (y – 2)2 ≤ 1

So, this indicates the area enclosed by a circle centred at (3, 2) with radius 1.

Now, the area of the region formed by x2 + y2 – 6x – 4y + 12 ≤ 0, y ≤ x and x ≤ 5/2 is

So, our bounds are x = 2 to x = 2.5

The equation for the ordinate of point on the circle from x = 2 to x = 2.5 –

(x – 3)2 + (y – 2)2 = 1

(y – 2)2 = 1 – (x – 3)2

Since we are considering the lower value of y in x = 2 to x = 2.5 (the one iny ≤ x),

y = 2 - \(\sqrt{1-(x-3)^2}\)

So, the area is –

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