Rewrite
in standard form find the center and radius of the circle.show all of your work for full credit​

Final answer:

The given equation can be rewritten into the form of an equation for a circle with center (h,k) and radius r. After completing the squares for x and y coefficients, we determine the center as (2, -5/2) and radius as 1.

Explanation:

The provided equation is 2x2 + 2y2 - 8x + 10y + 2 = 0.Let's rewrite it in the form of an equation for a circle, which is (x-h)^2 + (y-k)^2 = r^2, where (h,k) represents the center of the circle and r represents the radius:

First, we group the terms related to x and y separately like this: 2(x2- 4x) + 2(y2 + 5y) +2 = 0.

Now, re-arrange the terms to complete the square: 2[(x-2)^2 - 4] + 2[(y+5/2)^2 - (5/2)^2] + 2 = 0.

Re-arrange again, this time to make it equate to zero: (x-2)^2 + (y+5/2)^2 = 1. Hence, the circle's center (h,k) is at (2, -5/2) and the radius (r) is the square root of 1, which is 1.

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Answer:

first group x's with x's and y's with y's

then complete the squra with x's and y's

2x^2-8x+2y^2+10y+2=0

2(x^2-4x)+2(y^2+5y)+2=0

take 1/2 of linear coeficient and square

-4/2=-2, (-2)^2=4

5/2=2.5, 2.5^2=6.25

add that and negative inside

2(x^2-4x+4-4)+2(y^2+5y+6.25-6.25)+2=0

factor perfect squares

2((x-2)^2-4)+2((y+2.5)^2-6.25)+2=0

distribute

2(x-2)^2-8+2(y+2.5)^2-12.5+2=0

2(x-2)^2+2(y+2.5)^2-18.5=0

add 18.5 both sides

2(x-2)^2+2(y+2.5)^2=18.5

divide both sides by 2

(x-2)^2+(y+2.5)^2=9.25

that is a circle center (2,-2.5) with radius √9.25