Find the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } - 2 x + 6 y - 6 = 0$$
A) center: (-3, 1); radius: 4
B) center: (3, -1); radius: 16
C) center: (-1, 3); radius: 16
D) center: (1, -3); radius: 4

Question

Quizplus · Accepted Answer

To find the center and radius of the circle given by the equation $x^2 + y^2 - 2x + 6y - 6 = 0$, we complete the square for both $x$ and $y$. The equation can be rearranged and grouped as $(x^2 - 2x) + (y^2 + 6y) = 6$. Completing the square for $x^2 - 2x$ involves adding and subtracting $(\frac{2}{2})^2 = 1$, and for $y^2 + 6y$ involves adding and subtracting $(\frac{6}{2})^2 = 9$. Thus, the equation becomes $x^2 - 2x + 1 + y^2 + 6y + 9 = 6 + 1 + 9$, which simplifies to $(x - 1)^2 + (y + 3)^2 = 16$. This shows the center is at $(1, -3)$ and the radius is $\sqrt{16} = 4$.

Find the Center and Radius of the Circle x2+y2−2x+6y−6=0x ^ { 2 } + y ^ { 2 } - 2 x + 6 y - 6 = 0x2+y2−2x+6y−6=0

Find the Center and Radius of the Circle $x ^ { 2 } + y ^ { 2 } - 2 x + 6 y - 6 = 0$