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.
$x^{2}+y^{2}-2x - 6y - 2=0$
center: ( , )
radius:

Explanation:

Step1: Rewrite the equation by completing the square for x - terms

Group the x - terms and y - terms: $(x^{2}-2x)+(y^{2}-6y)=2$.
For the x - terms, $x^{2}-2x=(x - 1)^{2}-1$.

Step2: Rewrite the equation by completing the square for y - terms

For the y - terms, $y^{2}-6y=(y - 3)^{2}-9$.
The equation becomes $(x - 1)^{2}-1+(y - 3)^{2}-9=2$.

Step3: Simplify the equation to the standard form of a circle

Rearrange the equation: $(x - 1)^{2}+(y - 3)^{2}=2 + 1+9$.
So, $(x - 1)^{2}+(y - 3)^{2}=12$.
The standard form of a circle is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center and $r$ is the radius.

Answer:

Center: $(1,3)$
Radius: $\sqrt{12}=2\sqrt{3}\approx3.464$