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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?

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

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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 $r$.

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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).

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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.

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