$$x^2-20x+y^2-16y=-20$$

The equation above defines a circle in the $xy$-plane. What is the radius of the circle?

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We'll need to rewrite the equation into standard form:

$$(x-h)^2+(y-k)^2=r^2$$

In order to do this, we must complete the square as follows:

$$x^2-20x+y^2-16y=-20$$ $$x^2-20x+\colorbox{aqua}{$100$} +y^2-16y=-20+\colorbox{aqua}{$100$}$$ $$x^2-20x+100 +y^2-16y+\colorbox{yellow}{$64$}=-20+100+\colorbox{yellow}{$64$}$$ $$\underbrace{x^2-20x+100}_\text{perfect square trinomial} + \underbrace{y^2-16y+64}_\text{perfect square trinomials} = -20+100+64$$ $$(x-10)^2+(y-8)^2=144$$ $$(x-10)^2+(y-8)^2=12^2$$ $$r=\boxed{12}$$