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Find the equation of a circle which passes through $(4,-3)$ and $(-3,-4)$ with radius $5$.

I tried putting the $x$ and $y$ into the equation $(x-h)^2 + (y-k)^2 = r^2$, but then I don't know how to continue.

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  • $\begingroup$ If you know the famous 3-4-5 right-angled triangle, then you can see immediately that $(0,0)$ is one point which has distance 5 from the given points. $\endgroup$ – Martin Sleziak Jun 3 '15 at 14:45
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Since we have $$(4-h)^2+(-3-k)^2=5^2\iff h^2-8h+16+k^2+6k+9=25\tag1$$ $$(-3-h)^2+(-4-k)^2=5^2\iff h^2+6h+9+k^2+8k+16=25$$ subtracting the latter from the former gives you $$(-8-6)h+7+(6-8)k-7=0\iff k=-7h.$$ Then, use $(1)$ to get $h=0,1$. So, the answer is the followings : $$x^2+y^2=25,\ \ (x-1)^2+(y+7)^2=25.$$

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