Answer
The equation given represents a circle of center $\Big(\dfrac{3}{4},0\Big)$ and radius $\dfrac{3}{4}$
Work Step by Step
$2x^{2}+2y^{2}-3x=0$
Take out common factor $2$ from the left side of the equation:
$2\Big(x^{2}+y^{2}-\dfrac{3}{2}x\Big)=0$
Take $2$ to divide the right side:
$x^{2}+y^{2}-\dfrac{3}{2}x=\dfrac{0}{2}$
$x^{2}+y^{2}-\dfrac{3}{2}x=0$
Group the terms with $x$ together and group the terms with $y$ together:
$\Big(x^{2}-\dfrac{3}{2}x\Big)+y^{2}=0$
Complete the square for the first parentheses. Add $\Big(\dfrac{b}{2}\Big)^{2}$ inside the parentheses and do the same on the right side of the equation. In this particular case, $b=-\dfrac{3}{2}$.
$\Big[x^{2}-\dfrac{3}{2}x+\Big(-\dfrac{3}{2\cdot2}\Big)^{2}\Big]+y^{2}=\Big(-\dfrac{3}{2\cdot2}\Big)^{2}$
$\Big(x^{2}-\dfrac{3}{2}x+\dfrac{9}{16}\Big)+y^{2}=\dfrac{9}{16}$
Factor the expression inside the parentheses, which is a perfect square trinomial:
$\Big(x-\dfrac{3}{4}\Big)^{2}+y^{2}=\dfrac{9}{16}$
The expression obtained is the equation of a circle in $(x-h)^{2}+(y-k)^{2}=r^{2}$ form,
where $\Big(\dfrac{3}{4},0\Big)$ is the center of the circle and $\dfrac{3}{4}$ is its radius.
