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

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


y – 2 = y = 2


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


y = 2 -


So, the area is –




(Ans)

1