Three externally touching circles tangent to two paralle lines by HCR

Let the centers of circles $C_{1}, C_{2}, C_{3}$ be $P_{1}, P_{2}, P_{3}$ , respectively, and let the radius of $C_{3}$ be $r.$ We are thus looking for $2r.$

Draw lines $L_{1}$ and $L_{2}$ parallel to the given two parallel lines through $P_{1}$ and $P_{2}$ , respectively, and lines $V_{2}$ and $V_{3}$ perpendicular to the given two parallel lines through $P_{2}$ and $P_{3},$ respectively.

Now let $L_{1}$ intersect $V_{3}$ and $V_{2}$ at $A$ and $B$ , respectively, and $L_{2}$ intersect $V_{3}$ and $V_{2}$ at $D$ and $E$ , respectively.

Then $\Delta AP_{1}P_{3}$ is a right triangle with hypotenuse $P_{1}P_{3}$ length $6 + r$ and vertical leg $AP_{3}$ length $r - 6$ . By Pythagoras we then have that horizontal leg $AP_{1}$ has length

$\sqrt{(6 + r)^{2} - (r - 6)^{2}} = \sqrt{24r}.$

Next, we see that $\Delta DEP_{3}$ is a right triangle with hypotenuse $EP_{3}$ length $12 + r$ and vertical leg $DP_{3}$ length $r - 12.$ By Pythagoras we then have that horizontal leg $DE$ has length

$\sqrt{(12 + r)^{2} + (r - 12)^{2}} = \sqrt{48r}.$

Finally, we see that $\Delta BEP_{1}$ is a right triangle with hypotenuse $EP_{1}$ length $18$ , vertical leg $BE$ length $2r - 18$ and horizontal leg $BP_{1}$ length

$|DE| - |AP_{1}| = \sqrt{48r} - \sqrt{24r} = (\sqrt{2} - 1)\sqrt{24r}.$

By Pythagoras, we then have that

$18^{2} = (2r - 18)^{2} + (\sqrt{2} - 1)^{2}(24r)$

$\Longrightarrow 18^{2} = 4r^{2} - 72r + 18^{2} + (3 - 2\sqrt{2})(24r)$

$\Longrightarrow 4r^{2} - 48\sqrt{2}r = 0 \Longrightarrow 4r(r - 12\sqrt{2}) = 0.$

Now since $r \ne 0$ we see that $r = 12\sqrt{2}$ , and thus the distance between the original two parallel lines is $24\sqrt{2}$ , giving us $a + b = 24 + 2 = \boxed{26}.$

In general, if $r_1 \ and \ r_2$ are radii of small circles $C_1$ & $C_2$ then the radius $R$ of largest circle $C_3$ inscribed in a rectangle , is given by the general formula

$\boxed{R=2\sqrt{r_1 r_2}}$

As per given problem, setting $r_1=6, r_2=12$ the radius of largest circle is given as $R=2\sqrt{6\cdot 12}=2\sqrt{72}=12\sqrt2$

hence the distance between the parallel lines $2R=2\cdot12\sqrt2=24\sqrt2=a\sqrt b$

$\therefore a+b=24+2=26$