The real numbers $x$ and $y$ satisfy the equation $x^2+y^2=14x+48y-8x-26y+16$. What is the minimum value of $y$?
+4 First, x^2+y^2=14x+48y-8x-26y+16 rearrange as x^2 -6x + y^2 -22y = 16
complete the square on x, y
x^2 - 6x + 9 + y^2 -22y + 121 = 16 + 9 + 121
(x - 3)^2 + (y -11)^2 = 146
This is a circle centered at (3, 11) with a radius of sqrt (146)
The minimum value of y = 11 - sqrt (146) ≈ -1.083
