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?

• There is nothing you need to do except checking. See below. – Mick Nov 24 '14 at 3:40

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

• Thanks. I would vote up. I just realized this. Thanks for confirming it. – user194733 Nov 23 '14 at 0:41

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

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.

you can write it as $y^2+x^2-2\cdot \frac{3}{2}x+\frac{9}{4}=\frac{9}{4}$ this is equivalent to $y^2+\left(x-\frac{3}{2}\right)^2=\frac{9}{4}$

An equation for a circle (any circle) takes the general form:- $x^2 + y^2 + Dx + Ey + F = 0$.

The characteristics of it is (1) $((x^2)) = ((y^2))$ and (2) ((xy)) = 0; where ((t)) represents the coefficient of the term t.

The equation $y^2 + x^2 = 3x$ meets all of the the above criteria and therefore automatically is an equation of a circle.