Another Challenging Circle Problem

BD is minimum when it is perpendicular to AC.

$BD_{min}=4.8$

Also, as AC is tangent, and BD is perpendicular to it $\Rightarrow$ BD is diameter of circle.

According to given sides of triangle, angle B = 90°

So, MN is also diameter of circle (diameter subtends 90° angle at circumference)

So, $MN_{min}=BD_{min}=\boxed{4.8}$

Note that $\triangle ABC$ is a right triangle because $6^2+8^2=10^2$ , where $\angle B = 90^\circ$ . Since $B$ , $M$ , and $N$ are points on the circle, and $\angle B=90^\circ$ , $MN$ must be the diameter of the circle. The circle with the smallest diameter and hence the shortest $MN$ is when $BD$ is a straight line. Then $BD = MN$ , the diameter of the circle. And that $\triangle BCD$ is similar to $\triangle ABC$ , therefore $\frac {BD}{BC} = \frac {AB}{AC} = \frac 45$ $\implies BD = MN = \frac 45 BC = \frac 45 \times 6 = \boxed{4.8}$ .