To find the center of the circle given by the equation \( x^2 + y^2 + 2x - 20y + 5 = 0 \), we need to complete the square for both the \( x \) and \( y \) terms.
First, rearrange the equation:
\[ x^2 + 2x + y^2 - 20y + 5 = 0 \]
Now, let's complete the square for \( x \) and \( y \).
Completing the square for \( x \):
The terms we have are \( x^2 + 2x \).
- Take the coefficient of \( x \) which is \( 2 \), halve it to get \( 1 \), and square it to get \( 1 \).
- Thus, we can write:
\[ x^2 + 2x = (x + 1)^2 - 1 \]
Completing the square for \( y \):
The terms we have are \( y^2 - 20y \).
- Take the coefficient of \( y \) which is \( -20 \), halve it to get \( -10 \), and square it to get \( 100 \).
- Thus, we can write:
\[ y^2 - 20y = (y - 10)^2 - 100 \]
Substituting back into the equation:
Now substitute these completed squares back into the equation:
\[ ((x + 1)^2 - 1) + ((y - 10)^2 - 100) + 5 = 0 \]
Simplifying this gives:
\[ (x + 1)^2 + (y - 10)^2 - 1 - 100 + 5 = 0 \]
Combine like terms:
\[ (x + 1)^2 + (y - 10)^2 - 96 = 0 \]
Add \( 96 \) to both sides:
\[ (x + 1)^2 + (y - 10)^2 = 96 \]
Finding the center:
The standard form of a circle's equation is \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the center of the circle. From our equation, we see:
- \( h = -1 \)
- \( k = 10 \)
Thus, the center of the circle is:
\[ \boxed{(-1, 10)} \]