Find the Center and Radius 2x^2+2y^2+8x+7=0

Problem

2*x^2+2*y^2+8*x+7=0

Solution

1. Divide the entire equation by 2 to make the coefficients of x^2 and y^2 equal to 1

x^2+y^2+4*x+7/2=0

2. Rearrange the terms to group the x variables together and move the constant to the right side of the equation.

(x^2+4*x)+y^2=−7/2

3. Complete the square for the x terms by adding (4/2)^2=4 to both sides of the equation.

(x^2+4*x+4)+y^2=−7/2+4

4. Simplify the right side by finding a common denominator and factor the quadratic expression on the left side.

(x+2)^2+y^2=−7/2+8/2

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

5. Identify the center (h,k) and the radius r by comparing the result to the standard form of a circle (x−h)^2+(y−k)^2=r^2

h=−2

k=0

r^2=1/2

6. Calculate the radius by taking the square root of r^2 and rationalizing the denominator.

r=√(,1/2)

r=1/√(,2)

r=√(,2)/2

Final Answer

Center: *(−2,0), Radius: √(,2)/2

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