# Find the area of

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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