Equation of circle in quarter/half of a circle

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The equation for a quarter circle, y = ±√(r² - x²), represents both the upper and lower halves of the circle, but they are defined by different restrictions on x. For a quarter circle, the positive and negative roots are used separately, with specific x-value limits for each quadrant. In contrast, the equation for a half circle considers both roots simultaneously, allowing for a broader range of x-values from -r to r. The distinction lies in the domain restrictions applied to the x-values for each segment of the circle. Understanding these restrictions clarifies why the equations appear similar yet represent different geometric shapes.
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Homework Statement



I am curious why is the equation of a quarter of a circle (y = \pm \sqrt{r^{2}-x^2}) the same as half a circle? Shouldnt they be different?

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If you want quarter circle you have to use either + (for positive quarter) and - (for negative quarter) in the equation you mentioned.
For getting half circle both + and - ve values of equation are considered.
 
TsAmE said:

Homework Statement



I am curious why is the equation of a quarter of a circle (y = \pm \sqrt{r^{2}-x^2}) the same as half a circle? Shouldnt they be different?
Actually, they are different if you include restrictions on x. For example, the equation for the upper right quarter circle is
y = +\sqrt{r^{2}-x^2}, 0 \le x \le r

The equation for the upper left quarter circle has a different restriction on x; namely
-r \le x \le 0

For the upper half of the circle, you have -r \le x \le r

For the lower half circle and quarter circles, the only difference is that the negative square root is used.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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