Find the Center and Radius x^2+y^2+12x+2y+28=0

Problem

x2+y2+12*x+2*y+28=0

Solution

1. Group the x terms and y terms together and move the constant term to the right side of the equation.

x2+12*x+y2+2*y=−28

2. Complete the square for the x terms by adding (12/2)2=36 to both sides.

x2+12*x+36+y2+2*y=−28+36

3. Complete the square for the y terms by adding (2/2)2=1 to both sides.

x2+12*x+36+y2+2*y+1=−28+36+1

4. Factor the perfect square trinomials on the left and simplify the constant on the right.

(x+6)2+(y+1)2=9

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

h=−6

k=−1

r2=9⇒r=3

Final Answer

Center: *(−6,−1), Radius: *3

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