Question

Calculate the center and radius of the circle.

x2+y2+8x−6y−11=0 (1 point)
Responses

Center: (8, −6), Radius: 3
Center: (8, −6), Radius: 3

Center: (4, −3), Radius: 6
Center: (4, −3), Radius: 6

Center: (−8, 6), Radius: 3
Center: (−8, 6), Radius: 3

Center: (−4, 3), Radius: 6

Answer 1

To find the center and radius of the circle given by the equation:

\[ x^2 + y^2 + 8x - 6y - 11 = 0, \]

we first rearrange it into the standard form of a circle's equation:

\[ (x - h)^2 + (y - k)^2 = r^2, \]

where \((h, k)\) is the center and \(r\) is the radius.

1. **Group the \(x\) and \(y\) terms**:
\[
x^2 + 8x + y^2 - 6y = 11.
\]

2. **Complete the square for the \(x\) terms**:
\[
x^2 + 8x \rightarrow (x + 4)^2 - 16.
\]

3. **Complete the square for the \(y\) terms**:
\[
y^2 - 6y \rightarrow (y - 3)^2 - 9.
\]

4. **Substituting back into the equation**:
\[
(x + 4)^2 - 16 + (y - 3)^2 - 9 = 11.
\]
Simplifying this gives:
\[
(x + 4)^2 + (y - 3)^2 - 25 = 11.
\]
\[
(x + 4)^2 + (y - 3)^2 = 36.
\]

5. **Identify the center and radius**:
- The standard form indicates that the center \((h, k)\) is \((-4, 3)\) and the radius \(r\) is \(\sqrt{36} = 6\).

Therefore, the final answer is:

**Center: (-4, 3), Radius: 6**.