Question
A circle is drawn on a coordinate plane with the center point at (-8, -2) and a radius of 2. derive the equation of the circle in standard form, using the given information and the pythagorean theorem
Answers
GPT-4o mini
To derive the equation of a circle in standard form, we start with the general formula for a circle with center \((h, k)\) and radius \(r\):
\[
(x - h)^2 + (y - k)^2 = r^2
\]
In your case, the center of the circle is at the point \((-8, -2)\), which means \(h = -8\) and \(k = -2\). The radius \(r\) is given as \(2\).
Now, we can substitute the values into the equation:
1. Substitute \(h\) and \(k\):
\[
(x - (-8))^2 + (y - (-2))^2 = r^2
\]
This simplifies to:
\[
(x + 8)^2 + (y + 2)^2 = r^2
\]
2. Substitute \(r = 2\):
\[
(x + 8)^2 + (y + 2)^2 = 2^2
\]
3. Calculate \(2^2\):
\[
(x + 8)^2 + (y + 2)^2 = 4
\]
So the equation of the circle in standard form is:
\[
(x + 8)^2 + (y + 2)^2 = 4
\]
\[
(x - h)^2 + (y - k)^2 = r^2
\]
In your case, the center of the circle is at the point \((-8, -2)\), which means \(h = -8\) and \(k = -2\). The radius \(r\) is given as \(2\).
Now, we can substitute the values into the equation:
1. Substitute \(h\) and \(k\):
\[
(x - (-8))^2 + (y - (-2))^2 = r^2
\]
This simplifies to:
\[
(x + 8)^2 + (y + 2)^2 = r^2
\]
2. Substitute \(r = 2\):
\[
(x + 8)^2 + (y + 2)^2 = 2^2
\]
3. Calculate \(2^2\):
\[
(x + 8)^2 + (y + 2)^2 = 4
\]
So the equation of the circle in standard form is:
\[
(x + 8)^2 + (y + 2)^2 = 4
\]