Question

what is the center of the circle with equation $x^{2}+y^{2}-10x + 6y-2 = 0$?\
\\(\bigcirc\\) a. \\((0,6)\\)\
\\(\bigcirc\\) b. \\((-5,3)\\)\
\\(\bigcirc\\) c. \\((5,-3)\\)\
\\(\bigcirc\\) d. \\((5,3)\\)\
\\(\bigcirc\\) e. \\((6,0)\\)

Explanation:

Step1: Rearrange equation terms

$x^2 - 10x + y^2 + 6y = 2$

Step2: Complete the square for x

Take half of -10: $-5$, square it: $25$. Add to both sides.
$x^2 -10x +25 + y^2 +6y = 2 +25$

Step3: Complete the square for y

Take half of 6: $3$, square it: $9$. Add to both sides.
$x^2 -10x +25 + y^2 +6y +9 = 27 +9$

Step4: Rewrite in circle standard form

Standard form: $(x-h)^2+(y-k)^2=r^2$, where $(h,k)$ is center.
$(x-5)^2 + (y+3)^2 = 36$

Step5: Identify center coordinates

From $(x-5)^2 + (y-(-3))^2 = 36$, $h=5$, $k=-3$.

Answer:

C. (5, -3)