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# A circle is centered at and has a radius of units. Point is on the circle, has integer coordinates, and the value of the

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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?

Apr 25, 2019

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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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Apr 25, 2019