Question
A circle in the xy-plane has a center at (-7, -6) and a radius that is sqrt 13 units long. Which of the following is an equation of the circle?
Answers
GPT-4o mini
The standard form of the equation of a circle in the xy-plane with center \((h, k)\) and radius \(r\) is given by:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
In this case, the center of the circle is \((-7, -6)\), which means \(h = -7\) and \(k = -6\). The radius is given as \(\sqrt{13}\), so \(r = \sqrt{13}\).
First, we need to compute \(r^2\):
\[
r^2 = (\sqrt{13})^2 = 13
\]
Now we can substitute the values for \(h\), \(k\), and \(r^2\) into the standard equation of the circle:
\[
(x - (-7))^2 + (y - (-6))^2 = 13
\]
This simplifies to:
\[
(x + 7)^2 + (y + 6)^2 = 13
\]
Therefore, the equation of the circle is:
\[
(x + 7)^2 + (y + 6)^2 = 13
\]
\[
(x - h)^2 + (y - k)^2 = r^2
\]
In this case, the center of the circle is \((-7, -6)\), which means \(h = -7\) and \(k = -6\). The radius is given as \(\sqrt{13}\), so \(r = \sqrt{13}\).
First, we need to compute \(r^2\):
\[
r^2 = (\sqrt{13})^2 = 13
\]
Now we can substitute the values for \(h\), \(k\), and \(r^2\) into the standard equation of the circle:
\[
(x - (-7))^2 + (y - (-6))^2 = 13
\]
This simplifies to:
\[
(x + 7)^2 + (y + 6)^2 = 13
\]
Therefore, the equation of the circle is:
\[
(x + 7)^2 + (y + 6)^2 = 13
\]