Question
Which equation is the standard form of the equation of a circle drawn on the coordinate plane, derived from the Pythagorean theorem? The radius is 4 and the center point is (2,-2).
(X-2)^2+(y-2)^2=4
(X-6)^2+(y-2)^2=16
(X-2)^2+(y+2)^2=16
(X+2)^2+(y-6)^2=4
(X-2)^2+(y-2)^2=4
(X-6)^2+(y-2)^2=16
(X-2)^2+(y+2)^2=16
(X+2)^2+(y-6)^2=4
Answers
GPT-4o mini
The standard form of the equation of a circle with center \((h, k)\) and radius \(r\) is given by the formula:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
In your case, the center of the circle is \((2, -2)\) and the radius is \(4\). Therefore, we can substitute \(h = 2\), \(k = -2\), and \(r = 4\) into the formula.
1. The center coordinates give us \(h = 2\) and \(k = -2\).
2. The radius \(r = 4\), so \(r^2 = 16\).
Putting this all together, we get:
\[
(x - 2)^2 + (y + 2)^2 = 16
\]
Thus, the correct answer is:
\((X-2)^2+(y+2)^2=16\)
\[
(x - h)^2 + (y - k)^2 = r^2
\]
In your case, the center of the circle is \((2, -2)\) and the radius is \(4\). Therefore, we can substitute \(h = 2\), \(k = -2\), and \(r = 4\) into the formula.
1. The center coordinates give us \(h = 2\) and \(k = -2\).
2. The radius \(r = 4\), so \(r^2 = 16\).
Putting this all together, we get:
\[
(x - 2)^2 + (y + 2)^2 = 16
\]
Thus, the correct answer is:
\((X-2)^2+(y+2)^2=16\)