To find the center and radius from the general form \(x^2+y^2+2gx+2fy+c=0\), the center is \((-g, -f)\) and the radius is \(\sqrt{g^2+f^2-c}\). In this problem, \(2g=-6 \Rightarrow g=-3\), \(2f=-2 \Rightarrow f=-1\), \(c=-15\). Center is \((3,1)\), radius is \(\sqrt{9+1-(-15)} = \sqrt{25}=5\).
Step 1: Convert the circle's equation to standard form: \((x-h)^2 + (y-k)^2 = r^2\). This involves completing the square for both x and y terms. Given: \[ (x^2 - 6x) + (y^2 - 2y) = 15 \]Complete the square: add \((\frac{-6}{2})^2 = 9\) for x and \((\frac{-2}{2})^2 = 1\) for y.\[ (x^2 - 6x + 9) + (y^2 - 2y + 1) = 15 + 9 + 1 \]\[ (x-3)^2 + (y-1)^2 = 25 \]
Step 2: Identify the center \((\alpha, \beta)\) and radius \(\gamma\). From the standard form, the center is \((h, k) = (\alpha, \beta) = (3, 1)\). The radius squared is \(r^2 = \gamma^2 = 25\), thus the radius is \(\gamma = 5\).
Step 3: Calculate \(\alpha^2+\beta^2+\gamma^2\). \[ \alpha^2+\beta^2+\gamma^2 = (3)^2 + (1)^2 + (5)^2 \]\[ = 9 + 1 + 25 = 35 \]