To find the center and radius of the circle represented by the equation \( x^2 + y^2 - 10x + 4y = -13 \), we need to rewrite it in standard form. The standard form of a circle is given by
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius.
First, let's rearrange the given equation:
\[ x^2 - 10x + y^2 + 4y = -13 \]
Next, we will complete the square for both \(x\) and \(y\).
Completing the square for \(x\):
- Take the coefficient of \(x\), which is -10, halve it to get -5, and square it: \((-5)^2 = 25\).
- Add and subtract 25 in the equation:
\[ x^2 - 10x + 25 - 25 + y^2 + 4y = -13 \]
Completing the square for \(y\):
- Take the coefficient of \(y\), which is 4, halve it to get 2, and square it: \(2^2 = 4\).
- Add and subtract 4 in the equation:
\[ x^2 - 10x + 25 + y^2 + 4y + 4 - 25 - 4 = -13 \]
Now we write the equation as:
\[ (x - 5)^2 + (y + 2)^2 = -13 + 25 + 4 \]
Calculating the right side:
\[ -13 + 25 + 4 = 16 \]
So, we have:
\[ (x - 5)^2 + (y + 2)^2 = 16 \]
Now we can identify the center and radius:
- Center \((h, k) = (5, -2)\)
- Radius \(r = \sqrt{16} = 4\)
Thus, the answers are:
Center: (5, -2)
Radius: 4