Centre of circle I s(0, 0) Radius of circle = 4
`x^(2)+y^(2)=16" …(i)"`
`y=x" …(ii)"`
By equation (i) and (ii), we get
`x^(2)+x^(2)=16 rArr x^(2)=8`
`therefore" "x=2sqrt2.y=2sqrt2`
Shaded area = Area of OMPO + Area of PMQP
`=int_(0)^(2sqrt2)ydx+int_(2sqrt2)^(4)y dx`
by line by circle
`=int_(0)^(2sqrt2)xdx+int_(2sqrt2)^(4) sqrt(16-x^(2))dx`
`=[(x^(2))/(2)]_(0)^(2sqrt2)+[(x)/(2)sqrt(16-x^(2))+(16)/(2)sin^(-1).(x)/(4)]_(2sqrt2)^(4)`
`=((8)/(2)-0)+[0+8sin^(-1)1]-[(2sqrt2)/(2)sqrt(16-8)+8 sin^(-1).(1)/(sqrt2)]`
`=4+[8sin^(-1)1-sqrt2xx2sqrt2-8sin^(-1).(1)/(sqrt2)]`
`=4+(8xx(pi)/(2)-4-48xx(pi)/(4))`
`=4+4pi-4-2pi =2pi` square unit.
