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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?

 Feb 10, 2019
edited by Guest  Feb 10, 2019

Best Answer 

 #1
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\((2y-5)^2 + (y-15)^2 = 130\\ 4y^2 - 20y+25 + y^2 - 30y+225 = 130\\ 5y^2 - 50y +120=0\\ y^2 - 10y+24=0\\ (y-6)(y-4) = 0\\ y=6,~4\\ x=12,~8\\ max(x) = 12 \)

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 Feb 10, 2019
 #1
avatar+4419 
+1
Best Answer

\((2y-5)^2 + (y-15)^2 = 130\\ 4y^2 - 20y+25 + y^2 - 30y+225 = 130\\ 5y^2 - 50y +120=0\\ y^2 - 10y+24=0\\ (y-6)(y-4) = 0\\ y=6,~4\\ x=12,~8\\ max(x) = 12 \)

Rom Feb 10, 2019

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