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

Welcome back everyone. So in this problem, we're asked to determine if the equation x cubed, plus y squared, plus 4 x, minus 8 y, plus 4 equals 0 is a circle. And if it is a circle, to determine the center and radius of that circle. Now our first step when determining whether or not this is a circle is to check the general form of a circle since it appears we have that type of form. So So what I'm going to do, I'm gonna write down the general form of a circle which is x squared plus y squared plus ax plus by plus c is equal to 0. And what I need to do is look to see if we have a correct equation. But as you can see we have a problem here. This is an x cubed and our general form has an x squared. These forms of the equations do not match. And since they do not match, we can already determine that this is not a circle. And something else that I'll say is that if you thought this was a circle, and you went through solving this by trying to convert the standard form and find find the center and radius, you would find that this x cubed would actually give you an issue, because it would prevent you from being able to complete the square twice. So or you couldn't even complete it once with this these x terms. So this would actually cause a problem, meaning that you would find out it's actually not a circle. So that's how you solve this problem, that's the solution. Hope you found this helpful. Thanks for watching.

Determine if the equation x 3 + y 2 + 4 x − 8 y + 4 = 0 x^3+y^2+4x-8y+4=0 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 ( 0 , 0 ) c\left(0,0\right) c ( 0 , 0 ) , radius r = 4 r=4 r = 4 .

Is a circle, center = c ( 0 , 0 ) c ( 2 , − 4 ) c\left(0,0\right)c\left(2,-4\right) c ( 0 , 0 ) c ( 2 , − 4 ) , radius r = 4 r=4 r = 4 .

Is a circle, center = c ( − 2 , 4 ) c ( 0 , 0 ) c\left(-2,4\right)c\left(0,0\right) c ( − 2 , 4 ) c ( 0 , 0 ) , radius r = 4 r=4 r = 4 .

19. Conic Sections

Circles

Precalculus

19. Conic Sections

Circles

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Multiple Choice

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

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

Is a circle, center = $c\left(0,0\right)c\left(2,-4\right)$ , radius $r=4$ .

Is a circle, center = $c\left(-2,4\right)c\left(0,0\right)$ , radius $r=4$ .

Is not a circle.

Verified step by step guidance

Identify 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.

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

Notice that the term x^3 is present in the equation, which is not part of the standard form of a circle's equation. A circle's equation only includes x^2 and y^2 terms.

Since the equation includes an x^3 term, it cannot be rewritten in the form (x - h)^2 + (y - k)^2 = r^2.

Conclude that the given equation does not represent a circle because it does not fit the standard form of a circle's equation.

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Struggling with Precalculus?

Determine if the equation x3+y2+4x−8y+4=0x^3+y^2+4x-8y+4=0x3+y2+4x−8y+4=0 is a circle, and if it is, find its center and radius.

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