The equation of curves are y2 = 2x and x2 + y2 = 4x
Solving the equations,
![\[\begin{array}{*{35}{l}} {{x}^{2~}}-\text{ }4x\text{ }+\text{ }{{y}^{2~}}=\text{ }0 \\ {{x}^{2~}}-\text{ }4x\text{ }+\text{ }4\text{ }-\text{ }4\text{ }+\text{ }{{y}^{2~}}=\text{ }0 \\ {{\left( x\text{ }-\text{ }2 \right)}^{2}}~+\text{ }{{y}^{2}}~=\text{ }4 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-11c294c8862e39cc2e70d63f2b4e4c9e_l3.png "Rendered by QuickLaTeX.com")
It’s clearly seen that the equation of the circle having its centre (2, 0) and radius 2.
Solving x2 + y2 = 4x and y2 = 2x
x2 + 2x = 4x
![\[\begin{array}{*{35}{l}} {{x}^{2}}~+\text{ }2x\text{ }-\text{ }4x\text{ }=\text{ }0 \\ {{x}^{2}}~-\text{ }2x\text{ }=\text{ }0 \\ x\left( x\text{ }-\text{ }2 \right)\text{ }=\text{ }0 \\ So,\text{ }x\text{ }=\text{ }0,\text{ }2 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f78666d066902e1b0a5021e51d55bda8_l3.png "Rendered by QuickLaTeX.com")
Now, the area of the required region is given as
