A circle is centered at \((5,15)\) and has a radius of \(\sqrt{130}\) units. Point \(Q = (x,y)\) is on the circle, has integer coordinates, and the value of the x-coordinate is twice the value of the y-coordinate. What is the maximum possible value for x?
The form of the circle is
(x - 5)^2 + (y - 15)^2 = 130
Since the x coordinate is twice the y coordinate, we have
(2y - 5)^2 + (y - 15)^2 = 130 simplify
4y^2 - 20y + 25 + y^2 - 30y + 225 = 130
5y^2 - 50y + 120 = 0
y^2 - 10y + 24 = 0
(y - 6) (y - 4) = 0 set each factor to 0 and solve for y and we get that y = 6 and y = 4
So....the max value for x = 2y = 2(6) = 12
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