# Equation of a circle $x^2+y^2=r^2$

I'm working with an optimization problem, and faced a problem when trying to understand the area/curve in the problem, it is the following: $$x^2+y^2-y=0$$ I was trying to determine whether this area is compact or not, to find out that this function is a circle with radius $$\frac{1}{2}$$ and base in origin!

Now I know that the equation of a circle is: $$x^2 + y^2 = r^2$$

But I couldn’t really understand how $$\sqrt{y}$$ is considered as the radius $$\frac{1}{2}$$.

What is the best way to understand it?

• $(x-0)^{2}+(y-\frac 1 2)^{2}=\frac 1 4$ is circle with center $(0,\frac 1 2 )$ and radius $\frac 1 2$. Oct 6, 2022 at 6:42

## 3 Answers

You can solve it by using the general form of a circle: $$(x-a)^2 +(y-b)^2 = r^2$$. Since there is no $$x$$ in your equation, then $$a =0$$. You can use $$x^2+ (y-b)^2 =r^2$$ to find $$b$$ and $$r$$.

In regard to your other question as to whether the equation describes a compact curve, we might write it as $$x^2 \ = \ y - y^2 \ \ . \$$ For real numbers, the left side is always non-negative. In order for this to equal a possible value for the right side, we must have $$\ y - y^2 \ = \ y·(1 - y) \ \ \ge \ 0 \ \ . \$$ As this can only occur if both factors have the same sign, or if either factor equals zero, we must have $$\ 0 \ \le \ y \ \le \ 1 \ \ . \$$ But this means that $$0 \ \ \le \ y - y^2 \ \ \le \ \ \frac14 \ \ \Rightarrow \ \ 0 \ \ \le \ x^2 \ \ \le \ \ \frac14 \ \ \Rightarrow \ \ -\frac12 \ \ \le \ x \ \ \le \ \ \frac12 \ \ .$$ (It is simple enough to find the absolute maximum of $$\ y - y^2 \ \ . \ )$$

So the curve is bounded and a parameterization such as $$\ x \ = \ \frac12· \cos t \ \ , \ \ y \ = \ \frac12 + \frac12·\sin t \ \$$ will show that the curve is closed (and simple).

This is more an extended hint than a full answer, as you will learn more by working out the example yourself.

You need to "complete the square" [in $$y$$] by adding a constant to each side to make the left-hand the sum of $$x^2$$ plus something else squared.

You should find that the radius is $$\frac 12$$ as advertised. The centre can also be identified from this form.