The equations of the given lines are
y = x (i)
and x2 + y2 = 32 (ii)
from (i) and (ii), we get that the line and circle intersect each other at B(4,4) in the first quadrant.
Draw perpendicular BM to x-axis.
Therefore, the required area = area of the region OBMO + area of the region BMAB.
Now, the area of the region OBMO =
The area of the region BMAB =
=8π –(8+4π) =4π -8
Thus, the required area = area of the region OBMO + area of the region BMAB.
= 8 + 4π -8 = 4π sq units.
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