The real numbers $x$ and $y$ satisfy the equation $x^2+y^2=14x+48y-8x-26y+16$. What is the minimum value of $y$?
Combining like terms, the equation is $x^2 + y^2 = 6x + 22y + 16$. Completing the square in $x$ and $y$, we get $(x - 3)^2 + \left( y - \frac{11}{2} \right)^2 = \frac{25}{2}^2$, so $(x - 3)^2 + \left( y - \frac{11}{2} \right)^2 \ge \left( \frac{25}{2} \right)^2$. Thus, $(x - 3)^2 + \left( y - \frac{11}{2} \right)^2 \ge \frac{25}{2}^2 = \frac{625}{4}$.
This equation represents a circle centered at $(3,11/2)$ with radius $25/2$, and we want to find the lines of the form $y = k$ that are tangent to this circle and below it. Let $y = k$ be a tangent line that is below the circle. Then
\begin{align*}
(x - 3)^2 + \left( k - \frac{11}{2} \right)^2 &> \left( \frac{25}{2} \right)^2 \\
(x - 3)^2 + k^2 - 11k + \frac{121}{4} &> \frac{625}{4} \\
(x - 3)^2 + k^2 - 11k - \frac{504}{4} &> 0 \\
(x - 3)^2 + k^2 - 11k - 126 &> 0.
\end{align*}Hence, the $y$-coordinate of the point $(x - 3)^2 + \left( y - \frac{11}{2} \right)^2$ is largest on the line $x = 3$ when we are in the region below the circle and above $y = k$. Then
\begin{align*}
x^2 + \left( k - \frac{11}{2} \right)^2 &= (x - 3)^2 + \left( k - \frac{11}{2} \right)^2 \\
&\ge \frac{625}{4},
\end{align*}where equality occurs when $x = 3$. Hence, we want $\left( k - \frac{11}{2} \right)^2 \ge \frac{625}{4}$.
Since we want $\left( k - \frac{11}{2} \right)^2$ to be as small as possible, this tells us that $k = \frac{11}{2} + \frac{25}{2} = 18$. Hence, the minimum value of $y$ is $\boxed{18}$.
To find the minimum value of $y$, we need to rewrite the given equation in terms of $y$ only.
Starting with the equation $x^2+y^2=14x+48y-8x-26y+16$, let's group like terms:
$x^2 + y^2 -14x - 48y + 8x + 26y = 16$
Combining like terms on both sides, we get:
$x^2 - 6x + y^2 - 22y = 16$
Now, we want to complete the square for the $y$ terms. To do this, we need to add and subtract the square of half the coefficient of the $y$ terms (in this case, the coefficient is $-22$).
Adding and subtracting $(-22/2)^2 = 121$ within the parentheses:
$x^2 - 6x + y^2 - 22y + 121 - 121 = 16$
Notice that $x^2 - 6x + 121$ can be written as $(x-3)^2$. Similarly, $y^2 - 22y + 121$ can be written as $(y-11)^2$.
Substituting these expressions back into the equation, we have:
$(x-3)^2 + (y-11)^2 - 121 = 16$
Rearranging terms, we get:
$(x-3)^2 + (y-11)^2 = 16 + 121$
$(x-3)^2 + (y-11)^2 = 137$
This equation represents a circle centered at $(3, 11)$ with a radius of $\sqrt{137}$.
The minimum value of $y$ occurs at the bottommost point of the circle. Since the center of the circle is at $(3, 11)$ and the circle is symmetric, the minimum value of $y$ is $11 - \sqrt{137}$.