A farmer's house sits 2 miles due south of a river. The diagram below shows the path that the farmer takes from his house to the river everyday in order to bring water to a farm located 5 miles south from the river and 6 miles due east from the farmer's house. By choosing a different point $P$ on the river to draw water, the farmer is able to walk different distances to do this task. Find the minimum distance possible that the farmer has to walk.
(You may enter an exact answer or round to the nearest hundredth of a mile.)

Step 1 ：Let the distance from the farmer's house to the point P on the river be denoted as $x$, and the distance from point P to the farm be denoted as $y$.

Step 2 ：The farmer's house, the point P on the river, and the farm form a right triangle, with the distance from the farmer's house to the farm as the hypotenuse.

Step 3 ：By the Pythagorean theorem, we have: $y^2=(5+x{)}^2+6^2$.

Step 4 ：Solving for $y$, we get: $y=\sqrt{(5+x{)}^2+36}$.

Step 5 ：The total distance the farmer has to walk is the sum of $x$ and $y$, which we denote as $D$: $D=x+y$.

Step 6 ：Substituting $y$ into the equation for $D$, we get: $D=x+\sqrt{(5+x{)}^2+36}$.

Step 7 ：To find the minimum distance, we need to find the derivative of $D$ with respect to $x$, set it equal to zero, and solve for $x$: $D^{\prime }=1+\frac{(5+x)}{\sqrt{(5+x{)}^2+36}}$.

Step 8 ：Setting $D^{\prime }$ equal to zero, we get: $1+\frac{(5+x)}{\sqrt{(5+x{)}^2+36}}=0$.

Step 9 ：Solving for $x$, we get: $x=-1-\sqrt{41}$.

Step 10 ：However, since $x$ represents a distance, it cannot be negative. Therefore, the minimum distance occurs when $x=0$.

Step 11 ：Substituting $x=0$ into the equation for $D$, we get: $D=0+\sqrt{(5+0{)}^2+36}$.

Step 12 ：Simplifying, we get: $D=\sqrt{25+36}$.

Step 13 ：Further simplifying, we get: $D=\sqrt{61}$.

Step 14 ：Therefore, the minimum distance the farmer has to walk is $\sqrt{61}$ miles, or approximately 7.81 miles.

Step 15 ：So, the final answer is $\boxed{{\displaystyle \sqrt{61}}}$ miles.