Find the Center and Radius 3x^2+3y^2-6x+12y=0

Problem

3*x2+3*y2−6*x+12*y=0

Solution

1. Divide the entire equation by 3 to simplify the coefficients of the x2 and y2 terms to 1

x2+y2−2*x+4*y=0

2. Group the x terms and y terms together to prepare for completing the square.

(x2−2*x)+(y2+4*y)=0

3. Complete the square for the x terms by adding ((−2)/2)2=1 and for the y terms by adding (4/2)2=4 to both sides of the equation.

(x2−2*x+1)+(y2+4*y+4)=0+1+4

4. Factor the perfect square trinomials into binomial squares and simplify the right side.

(x−1)2+(y+2)2=5

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=−2

r2=5⇒r=√(,5)

Final Answer

Center: *(1,−2), Radius: √(,5)

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