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

Is a circle, center = c ( 0 , 0 ) c\left(0,0\right) , radius r = 4 r=4 .

Is a circle, center = c ( 2 , − 4 ) c\left(2,-4\right) , radius r = 4 r=4 r = 4 .

Is a circle, center = c ( − 2 , 4 ) c\left(-2,4\right) , radius r = 4 r=4 r = 4 .

8. Conic Sections

Circles

College Algebra

8. Conic Sections

Circles

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Multiple Choice

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

Is a circle, center = $c of open paren 0 comma 0 close paren$ , radius $r equals 4$ .

Is a circle, center = $c of open paren 2 comma negative 4 close paren$ , radius $r=4$ .

Is a circle, center = $c of open paren negative 2 comma 4 close paren$ , radius $r=4$ .

Is not a circle.

Verified step by step guidance

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.

Step 2: Compare the given equation $x^3 + y^2 + 4x - 8y + 4 = 0$ with the general form of a circle's equation.

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.

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.

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.