Question

for the circle having the given equation, find each coordinate of the center exactly. also find radius either exactly with \sqrt()\ or by rounding to 3 decimal places.
$x^{2}+y^{2}-6x - 2y+2 = 0$
center: ( )
radius:
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Explanation:

Step1: Rewrite the equation in standard form

Complete the square for $x$ and $y$ terms.
The general equation of a circle is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center and $r$ is the radius.
For the $x$ - terms: $x^{2}-6x=(x - 3)^{2}-9$.
For the $y$ - terms: $y^{2}-2y=(y - 1)^{2}-1$.
The given equation $x^{2}+y^{2}-6x - 2y+2 = 0$ can be rewritten as $(x - 3)^{2}-9+(y - 1)^{2}-1+2=0$.

Step2: Simplify the equation

$(x - 3)^{2}+(y - 1)^{2}-9 - 1+2=0$.
$(x - 3)^{2}+(y - 1)^{2}=8$.

Answer:

Center: $(3,1)$
Radius: $\sqrt{8}\approx2.828$