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. ${(x-1)}^2+{(y+2)}^2=3^2$; Center: $(1,-2)$ ; Radius: $3$
• C. ${(x+1)}^2+{(y-2)}^2=3^2$; Center: $(-1,2)$; Radius: $9$
• D. ${(x-1)}^2+{(y+2)}^2=3^2$; Center: $(-1,2)$; 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 and move the constant to the other side: \((x^2 - 2x) + (y^2 + 4y) = 4\).
3. Complete the square for the \(x\) terms: take half of the coefficient of \(x\) (which is \(-2\)), square it, and add inside the parentheses. Do the same for the \(y\) terms: take half of \$4$, square it, and add inside the parentheses. Remember to add the same values to the right side to keep the equation balanced.
4. Rewrite the equation with the completed squares: \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
5. Identify the center \((h, k)\) from the completed square terms and find the radius \(r\) by taking the square root of the constant on the right side.