Let (x,y) be an ordered pair of real numbers that satisfies the equation x^2 + y^2 = 14x + 40y. What is the minimum value of x?
+2668 Rewrite this equation as: \(x^2-14x+y^2-40y=0\)
Completing the square, we have: \((x-7)^2+(y-20)^2=449\)
The minimum value for x is the center (7) - the radius.
The radius is the square root of the number on the right-hand side.
You should be able to take it from here!
