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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 \(a + b + c\).

 Jun 13, 2023
 #1
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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.

 Jun 13, 2023
 #2
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Random nonsense.

Guest Jun 14, 2023
 #3
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I agree, the answerer is impaired mentally.

Guest Jun 14, 2023
 #4
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The actual answer is -50.

 Jun 17, 2023

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