Which equation is the standard form of the equation of a circle drawn on the coordinate plane, derived from the Pythagorean Theorem?

Question

(1 point)
Responses

(x−2)2+(y+2)2=16
(
𝑥
−
2
)
2
+
(
𝑦
+
2
)
2
=
16
left parenthesis x minus 2 right parenthesis squared plus left parenthesis y plus 2 right parenthesis squared equals 16

(x−2)2+(y−2)2=4
(
𝑥
−
2
)
2
+
(
𝑦
−
2
)
2
=
4
left parenthesis x minus 2 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals 4

(x−6)2+(y−2)2=16
(
𝑥
−
6
)
2
+
(
𝑦
−
2
)
2
=
16
left parenthesis x minus 6 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals 16

(x+2)2+(y−6)2=4

Answer 1

The standard form of the equation of a circle in the coordinate plane, derived from the Pythagorean theorem, 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.

From the given responses, we can analyze them as follows:

1. \((x−2)^2+(y+2)^2=16\) has center \((2, -2)\) and radius \(4\) (since \(r^2 = 16\)).
2. \((x−2)^2+(y−2)^2=4\) has center \((2, 2)\) and radius \(2\) (since \(r^2 = 4\)).
3. \((x−6)^2+(y−2)^2=16\) has center \((6, 2)\) and radius \(4\) (since \(r^2 = 16\)).
4. \((x+2)^2+(y−6)^2=4\) has center \((-2, 6)\) and radius \(2\) (since \(r^2 = 4\)).

All of the equations listed are valid standard forms of the equation of a circle.

If you need to choose one specific equation, any of the options would be correct as they all represent circles. However, if you have further instructions on which one to select based on a particular context (like radius size, center position, or a specific condition), please provide that context. Otherwise, any of the equations can be considered as the standard form of a circle.