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?

FlyEaglesFly Apr 25, 2019

#1**+1 **

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

7

CPhill Apr 25, 2019