Picture this ....

Just a quick outline of one solution method....

The distance between the center $C$ of the unit circle and the point $T$ of tangency of the chord and the smaller circle is $1 - 2r$ . By symmetry, $T$ is the midpoint of the chord, so triangle $CTP$ is a right-angled triangle. By Pythagoras, we have that

$TP = \sqrt{1 - (1 - 2r)^{2}} = 2\sqrt{r - r^{2}}$ ,

and so the length of the chord is $4\sqrt{r - r^{2}}$ .

Again using Pythagoras, the lengths of sides $OP$ and $OQ$ are each

$\sqrt{r^{2} + (TP)^{2}} = \sqrt{r^{2} + 4(r - r^{2})} = \sqrt{4r - 3r^{2}}$ .

The perimeter of triangle $OTP$ is then $L = 4\sqrt{r - r^{2}} + 2\sqrt{4r - 3r^{2}}$ .

Setting the derivative $\dfrac{dL}{dr}$ equal to $0$ and then simplifying yields the equation

$(2r - 1)\sqrt{4r - 3r^{2}} = (2 - 3r)\sqrt{r - r^{2}}$ ,

which, after squaring both sides and further simplifying, yields the equation

$3r^{2} - 7r + 3 = 0 \Longrightarrow r = \dfrac{7 \pm \sqrt{13}}{6}$ .

Now since $r \lt 1$ we have $r = \dfrac{7 - \sqrt{13}}{6}$ . Thus $a = 7, b = 13, c = 6$ and $a + b + c = \boxed{26}$ .

Comment: Since the perimeter would go to $0$ as $r \rightarrow 0$ it should be clear that any critical point found using the above method will yield a maximum perimeter. An application of the second derivative test would be definitive, but not practically necessary.