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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

 

 

cool cool cool7

 Apr 25, 2019

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