2x^2+2y^2-3x+4y+2=0 find the radius of the circle

Question

2x^2+2y^2-3x+4y+2=0 find the radius of the circle

Answer 1

Divide every term by 2 to make "completing the square" a bit simpler.

x² + y² -3x/2 +2y + 1 = 0

Rearrange terms in preparation for completing the square (the goal is to get the equation in the form

(x-h)² + (y-k)² = r².

x² - 3x/2 + ? + y² + 2y + ? = -1

To complete the square for the first two terms (filling in the first question mark), take 1/2 of -3/2 and square it--that completes the square for "x"). Similarly fill in the second question mark by taking 1/2 of 2 and square the result.

x² - 3x/2 +9/16 + y² + 2y + 1 = -1 + 9/16 + 1 (note how the complete the square values were added to both sides of the equation.

So r² = 9/16, making r = 3/4.

The equation could be written as (x - 3/4)² + (y+ 1)² = 9/16 allowing identification of the center of the circle as (3/4, -1).

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Answer 2

x^2 + y^2 - (3/2)x + 2y + 1 = 0 <--- divides everything by 2

x^2 -(3/2)x + y^2 + 2y = -1

x^2 - (3/2)x + 9/16 + y^2 + 2y + 1 = -1 + 9/16 + 1 <--- completes the square

(x - 3/4)^2 + (y+1)^2 = 9/16

the center is at (3/4, -1) and the radius is 3/4