Determine if the equation x2+y2−2x+4y−4=0x^2+y^2-2x+4y-4=0x2+y2−2x+4y−4=0 is a circle, and if it is, find its center and radius.

Is a circle, center = c ( 1 , − 2 ) c\left(1,-2\right) c ( 1 , − 2 ) , radius r = 3 r=3 r = 3 .

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

Is a circle, center = c ( 1 , − 2 ) c\left(1,-2\right) c ( 1 , − 2 ) , radius r = 3 r=3 r = 3 .

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

Is a circle, center = c ( 0 , 0 ) c\left(0,0\right) c ( 0 , 0 ) , radius r = 3 r=3 r = 3 .

8. Conic Sections

Circles

College Algebra

8. Conic Sections

Circles

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Multiple Choice

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

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

Is a circle, center = $c\left(0,0\right)$ , radius $r=3$ .

Is a circle, center = $c\left(1,-2\right)$ , radius $r=3$ .

Is not a circle.

Verified step by step guidance

Start by rewriting the given equation: $ x^2 + y^2 - 2x + 4y - 4 = 0 $.

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.

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 $.

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 $.

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 $3$.

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