Find the minimum value of $\sqrt{x^2+y^2}$ if $5x+12y=60$. 
$\textbf{(A)}\ \frac{60}{13}\qquad \textbf{(B)}\ \frac{13}{5}\qquad \textbf{(C)}\ \frac{13}{12}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 0$
Solution 1
Let $x^2 + y^2 = r^2$, so $r = \sqrt{x^2 + y^2}$.  Thus, this problem is really finding the shortest distance from the origin to the line $5x + 12y = 60$.

From the graph, the shortest distance from the origin to the line is the [altitude](https://artofproblemsolving.com/wiki/index.php/Altitude) to the hypotenuse of the right triangle with legs $5$ and $12$.  The hypotenuse is $13$ and the area is $30$, so the altitude to the hypotenuse is $\frac{60}{13}$, which is answer choice $\boxed{\textbf{(A)}}$.

Solution 2
Solve for $y$ in the linear equation.
\[12y = 60 - 5x\]
\[y = 5 - \frac{5x}{12}\]
Substitute $y$ in $\sqrt{x^2+y^2}$.
\[\sqrt{x^2 + (5 - \frac{5x}{12})^2}\]
\[\sqrt{x^2 + 25 - \frac{25x}{6} + \frac{25x^2}{144}}\]
\[\sqrt{\frac{169x^2}{144} - \frac{25x}{6} + 25}\]
To find the minimum, find the vertex of the quadratic.  The x-value of the vertex is $\frac{25}{6} \cdot \frac{72}{169} = \frac{300}{169}$.  Thus, the minimum value is
\[\sqrt{\frac{169}{144} \cdot \frac{300^2}{169^2} - \frac{25}{6} \cdot \frac{300}{169} + 25}\]
\[\sqrt{\frac{10000}{16 \cdot 169} - \frac{1250}{169} + 25}\]
\[\sqrt{\frac{625}{169} - \frac{1250}{169} + \frac{4225}{169}}\]
\[\sqrt{\frac{3600}{169}}\]
\[\frac{60}{13}\]
The answer is $\boxed{\textbf{(A)}}$.

Solution 3
By Cauchy-Schwarz, 
\[(5^2 + 12^2)(x^2 + y^2) \geq (5x+12y)^2\]
Therefore: 
\[(13^2)(x^2 + y^2) \geq 60^2\]
\[13\sqrt{x^2 + y^2} \geq 60\]
Therefore:
\[\sqrt{x^2 + y^2} \geq \frac{60}{13}\]
Thus the answer is $\boxed{\textbf{(A)}}$.