Find the Center and Radius 2x^2+2y^2-4x=0

Problem

2*x2+2*y2−4*x=0

Solution

1. Divide the entire equation by 2 to make the coefficients of x2 and y2 equal to 1

x2+y2−2*x=0

2. Rearrange the terms to group the x variables together and the y variables together.

(x2−2*x)+y2=0

3. Complete the square for the x terms by adding and subtracting ((−2)/2)2=1 inside the parentheses.

(x2−2*x+1)−1+y2=0

4. Factor the perfect square trinomial and move the constant to the right side of the equation.

(x−1)2+y2=1

5. Identify the center (h,k) and the radius r by comparing the equation to the standard form (x−h)2+(y−k)2=r2

h=1

k=0

r2=1⇒r=1

Final Answer

Center: *(1,0), Radius: *1

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