Find the ratio of the areas of the portion between the circle `x^2 + y^2 = a^2` and straight line `x = a /2`

Explanation

Let the line `x=(a)/(2)` intersect the circle `x^(2)+y^(2)=a^(2)` at points P and Q. Therefore two parts of the circle divided by the line are PAQP and PBQP.

Now area of PAQP
` = 2 xx` area of PAMP
`=2xxint_(a//2)^(a) sqrt(a^(2)-x^(2))dx`
`=2[(x)/(2)sqrt(a^(2)-x^(2))+(a^(2))/(2)"sin"^(-1)(x)/(a)]_(a//2)^(a)`
`=2[(0+(a^(2))/(2)"sin"^(-1)1)-((a)/(4).(asqrt(3))/(2)+(a^(2))/(2)"sin"^(-1)(1)/(2))]`
`=2((a^(2)pi)/(4)-(a^(2)sqrt(3))/(8)-(a^(2)pi)/(12))`
`=(6a^(2)pi-3a^(2)sqrt(3)-2a^(2)pi)/(12)`
`=(a^(2)(4pi-3sqrt(3)))/(12)sq.` units.
and area of PBQP = Area of circle - Area of PAQP
`=pia^(2)-(a^(2)(4pi-3sqrt(3)))/(12)`
`=(12pia^(2)-4pia^(2)+3sqrt(3)a^(2))/(12)=(a^(2)(8pi+3sqrt(3)))/(12)`
` :. ` Area of PBQP : Area of PAQP
`=(a^(2)(8pi+3sqrt(3)))/(12) : (a^(2)(4pi-3sqrt(3)))/(12)`
` =(8pi+3sqrt(3)):(4pi-3sqrt(3))`.