Getting co-ordinates from equation of a circle

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I am given the question:

How many points with integer coordinates lie on the circumference of circle $x^2 + y^2 = 5^2$

I know the general equation of a circle so the radius is 5. How do I go about it now ?

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You know the given equation is obeyed by $3^2+4^2=5^2$. So you can choose $x=$ and $y=4$ or $x=4$ and $y=3$. Also note that if $x$ and $y$ obey the equation, any combination of $(\pm x,\pm y)$ also obeys the equation. So the possible solutions are $(3,4)$, $(-3,4)$, $(-3,-4)$, $(3,-4)$, $(4,3)$, $(4,-3)$, $(-4,-3)$, $(-4,3)$.

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We want to find the integer solutions of the equation $x^2+y^2=5^2$ .

Observe that we must have $-5\le x,y\le 5$ . WLOG assume $x\lt y.$

We get : $x^2+x^2\lt x^2+y^2\implies 2x^2 \lt 25 \implies x\le3$

Setting $x=0$ , we get $y = \pm5$.

Setting $x=\pm1$ , we get $y = \pm \sqrt{24}$ , which is not a solution.

Setting $x=\pm2$ , we get $y = \pm \sqrt{21}$ , which is not a solution .

Setting $x=\pm3$ , we get $y = \pm4$ .

Now similarly take the case of $y\lt x$ and find all coordinates.

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We know that $3^2+4^2=5^2$. that is all point $(x, y)=(±3, ±4)$ or $(x,y)=(±4, ±3)$ can be on circumference of circle with radius 5.Also four points on axis $(x, y)=(0, ±5)$ and $(x, y)=(±5, 0)$. Hence number of points with integer coordinates is 12; two on each quadrant, together 8 and four on axis.