A circle is drawn on a coordinate plane with the center point at (−1,3)

Question

and a radius of 5. Derive the equation of the circle in standard form, using the given information and the Pythagorean Theorem.(1 point)
Responses

(x−3)2+(y−1)2=25
left parenthesis x minus 3 right parenthesis squared plus left parenthesis y minus 1 right parenthesis squared equals 25

(x+3)2+(y−1)2=25
left parenthesis x plus 3 right parenthesis squared plus left parenthesis y minus 1 right parenthesis squared equals 25

(x+1)2+(y−3)2=25
left parenthesis x plus 1 right parenthesis squared plus left parenthesis y minus 3 right parenthesis squared equals 25

(x−1)2+(y−3)2=25

Answer 1

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 this case, the center of the circle is given as \((-1, 3)\) and the radius is \(5\).

1. We identify \(h = -1\) and \(k = 3\).
2. The radius \(r = 5\), so \(r^2 = 25\).

Now, we can plug these values into the standard form equation:

\[ (x - (-1))^2 + (y - 3)^2 = 25 \]

This simplifies to:

\[ (x + 1)^2 + (y - 3)^2 = 25 \]

Therefore, the correct answer is:

\[ \boxed{(x + 1)^2 + (y - 3)^2 = 25} \]