Mathematics, 28.06.2019 00:00 nsheikh2407
How do you solve this problem? what is the length of a radius of the circle represented by the equation x^2+y^2-4x+4y=0 ? will award brainliest for best explanation.
Answers: 1
![\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](/tpl/images/0025/1638/46752.png)
![\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}](/tpl/images/0025/1638/89179.png)
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