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 $\boxed{18}$.
The real numbers $x$ and $y$ satisfy the equation $x^2+y^2=14x+48y-8x-26y+16$. What is the minimum value of $y$?
1 answer