In $\triangle ABC, AB = 10~ AC = 8$ and $BC = 6$. Circle $P$ is the circle with smallest radius which passes through $C$ and is tangent to $AB$. Let $Q$ and $R$ be the points of intersection, distinct from $C$ , of circle $P$ with sides $AC$ and $BC$, respectively. The length of segment $QR$ is
$\textbf{(A) }4.75\qquad \textbf{(B) }4.8\qquad \textbf{(C) }5\qquad \textbf{(D) }4\sqrt{2}\qquad  \textbf{(E) }3\sqrt{3}$
We know that triangle $RCQ$ is similar to triangle $ABC$. We draw a line to point $D$ on hypotenuse $AB$ such that $\angle QDR$ is $90 ^\circ$ and that $RDQC$ is a rectangle. Since triangle $RCQ$ is similar to triangle $ABC$, let $RC$ be $4x$ and $RD/CQ$ be $3x$. Now we have line segment $AQ$ = $8-3x$, and line segment $RB$ = $6-4x$. Since $BD + DA = AB$, we use simple algebra and Pythagorean Theorem to get $\sqrt {(3x)^2 + (6-4x)^2}$ + $\sqrt {(4x)^2 + (8-3x)^2}$ = $10$. Expanding and simplifying gives us $\sqrt {25x^2-48x+36}$ + $\sqrt {25x^2-48x+64}$ = $10$. 
Squaring both sides is not an option since it is both messy and time consuming, so we should proceed to subtracting both sides by  $\sqrt {25x^2-48x+36}$. Now, we can square both sides and simplify to get $0 = 72 - 20 \sqrt{25x^2-48x+36}$. Dividing both sides by $4$, we get $18 - 5 \sqrt {25x^2-48x+36}$ = $0$. We then add $5 \sqrt {25x^2-48x+36}$ to both sides to get $18 = 5 \sqrt {25x^2-48x+36}$. Since this is very messy, let $25x^2 - 48x = y$. Squaring both sides, we get $324 = 25y + 900, 25y = -576$. Solving for $y$, we have $y = -23.04$. Plugging in $y$ as $25x^2-48x$, we have $25x^2-48x+23.04 = 0$. Using the quadratic equation, we get $\frac {48+0}{50}$. Therefore, $x = \frac {48}{50}$. 
Remember that our desired answer is the hypotenuse of the triangle $3x - 4x - 5x$. Since $5x$ is the hypotenuse, our answer is $\boxed {(B)4.8}$
~Arcticturn