Question 1156275: Let (x,y) be an ordered pair of real numbers that satisfies the equation
x^2+y^2=14x+48y. What is the minimum value of x? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! rewrite as x^2-14x+y^2-48y=0
complete the squares for each
x^2-14x+49+y^2-48y+576=625, adding 625 to the right
(x-7)^2+(y-24)^2=25^2
This is a circle with center (7, 24) and radius 25
the minimum value of x is (7-25)=-18
