Question

Calculate the center of the circle by completing the square of the equation x2+y2+2x−20y+5=0
.(1 point)
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,
)

Answer 1

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 \( x^2 + 2x \).
1. Take the coefficient of \( x \) which is \( 2 \), halve it to get \( 1 \), and square it to get \( 1 \).
2. Thus, we can write:

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

### Completing the square for \( y \):
The terms we have are \( y^2 - 20y \).
1. Take the coefficient of \( y \) which is \( -20 \), halve it to get \( -10 \), and square it to get \( 100 \).
2. 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:

\[
\boxed{(-1, 10)}
\]