+816 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 y?
+128407 x^2 + y^2 = 14x + 48y rearrange
x^2 - 14x + y^2 - 48y = 0 complete the square on x, y
(x^2 - 14x + 49) + (y^2 - 48y + 576) = 49 + 576
(x - 7)^2 + ( y - 24)^2 = 625
(x - 7)^2 + (y - 24)^2 = 25^2
This is a circle with a center at ( 7, 24) and a radius of 25
The minimum y value will occur at ( 7, 24 - 25) = ( 7, -1)
So...the minimum y value = -1
