Question

2x^{2}+2y^{2}+12x + 12y+28 = 0 is the equation of a circle with center (h,k) and radius r for: h = and k = and r =

Explanation:

Step1: Divide the equation by 2

Divide \(2x^{2}+2y^{2}+12x + 12y+28 = 0\) by 2 to get \(x^{2}+y^{2}+6x + 6y+14 = 0\).

Step2: Complete the square for \(x\) - terms

For \(x^{2}+6x\), we have \(x^{2}+6x=(x + 3)^{2}-9\).

Step3: Complete the square for \(y\) - terms

For \(y^{2}+6y\), we have \(y^{2}+6y=(y + 3)^{2}-9\).

Step4: Rewrite the equation

Substitute into the equation: \((x + 3)^{2}-9+(y + 3)^{2}-9+14 = 0\).
Simplify to \((x + 3)^{2}+(y + 3)^{2}=4\).

Answer:

\(h=-3\), \(k = - 3\), \(r = 2\)