There exists a circle \(C\) with radius \(5\), passing through the point \((3, 2)\), such that the center of C lies on the line \(x = 7\) and has positive \(y\)-coordinate. \(C\) is the graph of the equation \(x^2 + ax + y^2 + by + c = 0\), where \(a, b, c\) are integers (not necessarily positive). Find \(a + b + c\).
The center of the circle is (7,y). Since the circle passes through (3,2), we have the equation:
(7-3)^2 + a(7-3) + (y-2)^2 + by + c = 0
Expanding, we get:
16 + 4a + y^2 + by + c = 0
We are given that the radius of the circle is 5, so we can substitute this into the equation:
16 + 4a + y^2 + by + c = 25
Solving for a, b, and c, we get:
a = -3 b = -2 c = -3
Therefore, a + b + c = -8.
