Determine if the equation $x^3+y^2+4x-8y+4=0$ is a circle, and if it is, find its center and radius.

• A. Is a circle, center = $c(0,0)$, radius $r=4$.
• B. Is a circle, center = $c(2,-4)$ , radius $r=4$.
• C. Is a circle, center = $c(-2,4)$ , radius $r=4$.
• D. Is not a circle.

Answer: (D)

Verified step by step guidance

1. Step 1: Recall the general form of a circle's equation, which is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
2. Step 2: Compare the given equation \(x^3 + y^2 + 4x - 8y + 4 = 0\) with the general form of a circle's equation.
3. Step 3: Notice that the term \(x^3\) is present in the equation. In the equation of a circle, the highest power of \(x\) and \(y\) should be 2, indicating that the equation should only contain \(x^2\) and \(y^2\) terms.
4. Step 4: Since the equation contains an \(x^3\) term, it does not fit the standard form of a circle's equation, which means it cannot represent a circle.
5. Step 5: Conclude that the given equation is not a circle because it does not match the form \((x - h)^2 + (y - k)^2 = r^2\) due to the presence of the \(x^3\) term.