Circle radius as variable

I am confused. How is $y^2 + x^2 =3x$ a circle? Can someone please help me try to understand why the above a circle, or is it just a typo?

$$y^2+x^2 = 3x \implies y^2 + x^2-3x + (3/2)^2 = (3/2)^2 \implies (x-3/2)^2 + y^2 = (3/2)^2$$ This is a circle centered at $(3/2,0)$ with radius $3/2$.

Try completing the square on $x$ to see why it's a circle.

Complete the square: $$ x^2 - 3x = \left(x^2 - 3x + \frac 94\right) - \frac 9 4 = \left(x - \frac 3 2\right)^2 - \frac 9 4. $$ Therefore $y^2+x^2-3x=0$ is equivalent to $y^2 + \left(x - \dfrac 3 2 \right)^2 = \dfrac 9 4$.

The purpose of completing the square is always to reduce a problem involving a quadratic polynomial with a first-degree term to a problem involving a quadratic polynomial without a first-degree term.