Question
The equation of a circle is (x + 2)^2 + (y + 6)^2 = 16. What are the center and radius of the circle?
Answers
GPT-4o mini
To find the center and radius of the circle given by the equation \((x + 2)^2 + (y + 6)^2 = 16\), we can compare this equation with the standard form of a circle's equation, which is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
where \((h, k)\) is the center of the circle and \(r\) is the radius.
From the given equation \((x + 2)^2 + (y + 6)^2 = 16\):
1. Rewrite the equation in standard form:
- The term \((x + 2)\) can be rewritten as \((x - (-2))\), so \(h = -2\).
- The term \((y + 6)\) can be rewritten as \((y - (-6))\), so \(k = -6\).
- The right side, \(16\), represents \(r^2\), so \(r = \sqrt{16} = 4\).
Therefore, the center of the circle is \((-2, -6)\) and the radius is \(4\).
**Summary:**
- Center: \((-2, -6)\)
- Radius: \(4\)
\[
(x - h)^2 + (y - k)^2 = r^2
\]
where \((h, k)\) is the center of the circle and \(r\) is the radius.
From the given equation \((x + 2)^2 + (y + 6)^2 = 16\):
1. Rewrite the equation in standard form:
- The term \((x + 2)\) can be rewritten as \((x - (-2))\), so \(h = -2\).
- The term \((y + 6)\) can be rewritten as \((y - (-6))\), so \(k = -6\).
- The right side, \(16\), represents \(r^2\), so \(r = \sqrt{16} = 4\).
Therefore, the center of the circle is \((-2, -6)\) and the radius is \(4\).
**Summary:**
- Center: \((-2, -6)\)
- Radius: \(4\)