Question

The equation of a circle in general form is x squared plus y squared plus 2 x minus 8 y plus 16 equals 0.

What are the center and radius of the circle?
(1 point)
Responses

center: left-parenthesis negative 1 comma 4 right-parenthesis, r = 4
center: Image with alt text: left-parenthesis negative 1 comma 4 right-parenthesis , r = 4

center: left-parenthesis negative 1 comma negative 4 right-parenthesis, r = 4
center: Image with alt text: left-parenthesis negative 1 comma negative 4 right-parenthesis , r = 4

center: left-parenthesis negative 1 comma 4 right-parenthesis, r = 1
center: Image with alt text: left-parenthesis negative 1 comma 4 right-parenthesis , r = 1

center: left-parenthesis negative 1 comma negative 4 right-parenthesis, r = 1
center: Image with alt text: left-parenthesis negative 1 comma negative 4 right-parenthesis , r = 1

Answers

GPT-4o mini
To find the center and radius of the circle described by the equation \(x^2 + y^2 + 2x - 8y + 16 = 0\), we need to rewrite the equation in the standard form of a circle, which is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

We can start by rearranging the given equation:

1. Move the constant term to the other side of the equation:

\[
x^2 + y^2 + 2x - 8y + 16 = 0 \implies x^2 + y^2 + 2x - 8y = -16
\]

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

- For \(x^2 + 2x\):
- Take half of the coefficient of \(x\) (which is 2), square it: \(\left(\frac{2}{2}\right)^2 = 1\).
- Therefore, \(x^2 + 2x\) can be rewritten as \((x + 1)^2 - 1\).

- For \(y^2 - 8y\):
- Take half of the coefficient of \(y\) (which is -8), square it: \(\left(\frac{-8}{2}\right)^2 = 16\).
- Therefore, \(y^2 - 8y\) can be rewritten as \((y - 4)^2 - 16\).

3. Substitute these complete squares back into the equation:

\[
(x + 1)^2 - 1 + (y - 4)^2 - 16 = -16
\]

Simplifying gives:

\[
(x + 1)^2 + (y - 4)^2 - 17 = -16
\]

\[
(x + 1)^2 + (y - 4)^2 = 1
\]

4. Now the equation is in standard form \((x + 1)^2 + (y - 4)^2 = 1\).

From this equation, we can identify:

- The center \((h, k)\) is \((-1, 4)\).
- The radius \(r\) is given by \(\sqrt{1} = 1\).

Thus, the center and radius of the circle are:

**center: \((-1, 4)\), \(r = 1\)**.

The correct response is:
**center: \((-1, 4)\), r = 1**.