Question

Calculate the center and radius of the circle by completing the square of the following equation. x2+y2+8x−6y−11=0 (1 point) Responses Center: (−4, 3) , Radius: 6 Center: (−4, 3) , Radius: 6 Center: (4, −3) , Radius: 6 Center: (4, −3) , Radius: 6 Center: (8, −6) , Radius: 3 Center: (8, −6) , Radius: 3 Center: (−8, 6) , Radius: 3

Answer 1

To find the center and radius of the circle given by the equation \(x^2 + y^2 + 8x - 6y - 11 = 0\), we will complete the square for both \(x\) and \(y\).

First, rearrange the equation:

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

Next, we will complete the square for the \(x\) terms and the \(y\) terms.

1. **Completing the square for \(x^2 + 8x\)**:
- Take half of the coefficient of \(x\) (which is 8), square it: \((\frac{8}{2})^2 = 16\).
- Thus, \(x^2 + 8x\) can be rewritten as \((x + 4)^2 - 16\).

2. **Completing the square for \(y^2 - 6y\)**:
- Take half of the coefficient of \(y\) (which is -6), square it: \((\frac{-6}{2})^2 = 9\).
- Thus, \(y^2 - 6y\) can be rewritten as \((y - 3)^2 - 9\).

Now substitute these completed squares back into the equation:

\[
(x + 4)^2 - 16 + (y - 3)^2 - 9 = 11
\]

Combine like terms:

\[
(x + 4)^2 + (y - 3)^2 - 25 = 11
\]

Add 25 to both sides to isolate the completed squares:

\[
(x + 4)^2 + (y - 3)^2 = 36
\]

Now we can identify the center and radius of the circle from the equation \((x - h)^2 + (y - k)^2 = r^2\):

- The center \((h, k)\) is \((-4, 3)\).
- The radius \(r\) is \(\sqrt{36} = 6\).

Thus, the center and radius of the circle are:

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