Write the standard form equation of the circle described. Give the center and radius.
$x^2+y^2-2x+4y-4=0$

• A. ${(x+1)}^2+{(y-2)}^2=3^2$; Center: $(1,-2)$; Radius: $3$
• B. $\left(x-1\right)^2+\left(y+2\right)^2=3^2$; Center: $\left(1,-2\right)$; Radius: $3$
• C. $\left(x+1\right)^2+\left(y-2\right)^2=3^2$; Center: $\left(-1,2\right)$; Radius: $9$
• D. $\left(x-1\right)^2+\left(y+2\right)^2=3^2$; Center: $\left(-1,2\right)$; Radius: $3$

Answer: (B)

Verified step by step guidance

1. Start with the given general form of the circle equation: \(x^2 + y^2 - 2x + 4y - 4 = 0\).
2. Group the \(x\) terms and \(y\) terms together: \((x^2 - 2x) + (y^2 + 4y) = 4\) (move the constant to the right side).
3. Complete the square for the \(x\) terms: take half of the coefficient of \(x\) (which is \(-2\)), square it, and add it inside the parentheses. Do the same for the \(y\) terms: take half of \$4$, square it, and add it inside the parentheses.
4. Add the same values to the right side of the equation to keep it balanced. Then rewrite the expressions as perfect square trinomials: \((x - h)^2\) and \((y - k)^2\).
5. Write the equation in standard form: \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius (the square root of the number on the right side).