Determine if the equation $x^2+y^2-2x+4y-4=0$ is a circle, and if it is, find its center and radius.

• A. Is a circle, center = $c\left(0,0\right)$, radius $r=2$.
• B. Is a circle, center = $c\left(0,0\right)$, radius $r=3$ .
• C. Is a circle, center = $c\left(1,-2\right)$, radius $r=3$.
• D. Is not a circle.

Answer: (C)

Verified step by step guidance

1. Start by rewriting the given equation: \( x^2 + y^2 - 2x + 4y - 4 = 0 \).
2. To determine if this is a circle, we need to rewrite it in the standard form of a circle's equation: \( (x - h)^2 + (y - k)^2 = r^2 \), where \((h, k)\) is the center and \(r\) is the radius.
3. Complete the square for the \(x\) terms: \( x^2 - 2x \). Take half of the coefficient of \(x\), square it, and add it inside the equation: \( (x - 1)^2 - 1 \).
4. Complete the square for the \(y\) terms: \( y^2 + 4y \). Take half of the coefficient of \(y\), square it, and add it inside the equation: \( (y + 2)^2 - 4 \).
5. Rewrite the equation incorporating the completed squares: \( (x - 1)^2 + (y + 2)^2 = 9 \). This shows the equation is a circle with center \((1, -2)\) and radius \(3\).