Question

How do you solve this problem?

Answer 1

so, if you checked the link above, you know what we'll be doing, lemme run through it without much fuss.

$\bf \stackrel{\textit{firstly some grouping}}{(x^2-4x)+(y^2+4y)=0}\implies (x^2-4x+\boxed{a}^2)+(y^2+4y+\boxed{b}^2)=0 \\\\[-0.35em] ~\dotfill\\\\ 2\cdot x\cdot \boxed{a}=4x\implies \boxed{a}=\cfrac{4x}{2x}\implies \boxed{a}=2 \\\\\\ 2\cdot y\cdot \boxed{b}=4y\implies \boxed{b}=\cfrac{4y}{2y}\implies \boxed{b}=2$

now, let's recall, we're simply borrowing from our very good friend Mr Zero, 0, so if we add whatever, we also have to subtract whatever.

$\bf (x^2-4x+2^2-2^2)+(y^2+4y+2^2-2^2)=0 \\\\\\ (x^2-4x+2^2)+(y^2+4y+2^2)-4-4=0 \\\\\\ (x-2)^2+(y+2)^2-8=0\implies (x-2)^2+(y+2)^2=8 \\\\[-0.35em] ~\dotfill\\\\ \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r} \\\\[-0.35em] ~\dotfill\\\\ (x-2)^2+(y+2)^2=(\sqrt{8})^2\qquad \impliedby radius=\sqrt{8}$