Find the Center and Radius 2(x-3)^2+2y^2=8

$2 {(x - 3)}^{2} + 2 y^{2} = 8$

Step 1

Divide both sides of the equation by $2$ .

${(x - 3)}^{2} + y^{2} = 4$

Step 2

This is the form of a circle. Use this form to determine the center and radius of the circle.

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$

Step 3

Match the values in this circle to those of the standard form. The variable $r$ represents the radius of the circle, $h$ represents the x-offset from the origin, and $k$ represents the y-offset from origin.

$r = 2$

$h = 3$

$k = 0$

Step 4

The center of the circle is found at $(h, k)$ .

Center: $(3, 0)$

Step 5

These values represent the important values for graphing and analyzing a circle.

Center: $(3, 0)$

Radius: $2$

Step 6

Find the Center and Radius 2(x-3)^2+2y^2=8

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