Question

using the following equation, find the center and radius:
$x^{2}+2x+y^{2}+4y = 20$
the center is located at $(1,2)$, and the radius is 5.
the center is located at $(-1,-2)$, and the radius is 5.
the center is located at $(-1,2)$, and the radius is 25.
the center is located at $(1,2)$, and the radius is 25.

Explanation:

Step1: Complete the square for \(x\) terms

For \(x^{2}+2x\), we add \(1\) (since \((\frac{2}{2})^{2}=1\)).
\(x^{2}+2x=(x + 1)^{2}-1\)

Step2: Complete the square for \(y\) terms

For \(y^{2}+4y\), we add \(4\) (since \((\frac{4}{2})^{2}=4\)).
\(y^{2}+4y=(y + 2)^{2}-4\)

Step3: Rewrite the equation

The original equation \(x^{2}+2x+y^{2}+4y = 20\) becomes \((x + 1)^{2}-1+(y + 2)^{2}-4=20\).
Simplify to \((x + 1)^{2}+(y + 2)^{2}=20 + 1+4\).
So \((x + 1)^{2}+(y + 2)^{2}=25\).

The standard form of a circle equation is \((x - a)^{2}+(y - b)^{2}=r^{2}\), where \((a,b)\) is the center and \(r\) is the radius.

Here \(a=-1\), \(b = - 2\), \(r^{2}=25\) (so \(r = 5\))

Answer:

The center is located at \((-1,-2)\), and the radius is \(5\) (i.e., the second option).