Given:

The condition of the given circle is \[2{{x}^{2}}~+\text{ }2{{y}^{2}}~x\text{ }=\text{ }0.\]

\[\begin{array}{*{35}{l}}

2{{x}^{2}}~+\text{ }2{{y}^{2}}~x\text{ }=\text{ }0  \\

\left( 2{{x}^{2}}~+\text{ }x \right)\text{ }+\text{ }2{{y}^{2}}~=\text{ }0  \\

\left( {{x}^{2}}~\text{ }2\text{ }\left( x \right)\text{ }\left( 1/4 \right)\text{ }+\text{ }{{\left( 1/4 \right)}^{2}} \right)\text{ }+\text{ }{{y}^{2}}~\text{ }{{\left( 1/4 \right)}^{2}}~=\text{ }0  \\

{{\left( x\text{ }\text{ }1/4 \right)}^{2}}~+\text{ }{{\left( y\text{ }\text{ }0 \right)}^{2}}~=\text{ }{{\left( 1/4 \right)}^{2}}~\left[ which\text{ }is\text{ }form\text{ }{{\left( x-h \right)}^{2}}~+{{\left( y-k \right)}^{2}}~=\text{ }{{r}^{2}} \right]  \\

\end{array}\]

Where, \[~h\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_4},\text{ }K\text{ }=\text{ }0,\text{ }and\text{ }r\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\]

∴ The focal point of the given circle is \[\left( 1/4,\text{ }0 \right)~\]and its sweep is \[1/4.\]