Question
Which equation is the standard form of the equation of a circle drawn on the coordinate plane, derived from the Pythagorean Theorem? c is (2,-2) p is (6,-2) (1 point)Responses(x+2)2+(y−6)2=4left parenthesis x plus 2 right parenthesis squared plus left parenthesis y minus 6 right parenthesis squared equals 4(x−6)2+(y−2)2=16left parenthesis x minus 6 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals 16(x−2)2+(y−2)2=4left parenthesis x minus 2 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals 4(x−2)2+(y+2)2=16
Answers
GPT-4o mini
The equation of a circle in standard form is given by the formula:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
where \((h, k)\) is the center of the circle, and \(r\) is the radius.
In your case, the center \(c\) is \((2, -2)\) and the point \(p\) is \((6, -2)\). The radius \(r\) can be calculated using the distance between the center and the point, which can be found using the distance formula:
\[
r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Plugging in the coordinates of \(c\) and \(p\):
\[
r = \sqrt{(6 - 2)^2 + (-2 - (-2))^2} = \sqrt{(4)^2 + (0)^2} = \sqrt{16} = 4
\]
Now that we have the center \((2, -2)\) and the radius \(r = 4\), we can write the standard form of the equation of the circle:
\[
(x - 2)^2 + (y + 2)^2 = 4^2
\]
\[
(x - 2)^2 + (y + 2)^2 = 16
\]
Thus, the correct standard form of the equation of the circle is:
\[
\boxed{(x - 2)^2 + (y + 2)^2 = 16}
\]
\[
(x - h)^2 + (y - k)^2 = r^2
\]
where \((h, k)\) is the center of the circle, and \(r\) is the radius.
In your case, the center \(c\) is \((2, -2)\) and the point \(p\) is \((6, -2)\). The radius \(r\) can be calculated using the distance between the center and the point, which can be found using the distance formula:
\[
r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Plugging in the coordinates of \(c\) and \(p\):
\[
r = \sqrt{(6 - 2)^2 + (-2 - (-2))^2} = \sqrt{(4)^2 + (0)^2} = \sqrt{16} = 4
\]
Now that we have the center \((2, -2)\) and the radius \(r = 4\), we can write the standard form of the equation of the circle:
\[
(x - 2)^2 + (y + 2)^2 = 4^2
\]
\[
(x - 2)^2 + (y + 2)^2 = 16
\]
Thus, the correct standard form of the equation of the circle is:
\[
\boxed{(x - 2)^2 + (y + 2)^2 = 16}
\]