Circumferences

Let $R$ be the radius of the larger circle, and let $M$ be the center of the smaller circle. Then

$(MQ)^{2} = (OQ)^{2} + (MO)^{2}$ .

Now $(OQ) = R - 3$ ,

$(MQ) = (MP) = \frac{1}{2}(2R - 4) = R - 2$ , and

$(MO) = (MP) - (OP) = (R - 2) - (R - 4) = 2$ .

So we have that

$(R - 2)^{2} = (R - 3)^{2} + 2^{2} \Longrightarrow R^{2} - 4R + 4 = R^{2} - 6R + 9 + 4$

$\Longrightarrow 2R = 9 \Longrightarrow R = 4.5$ .

So the radius of the smaller circle is $(MQ) = R - 2 = \boxed{2.5}$ .

We can apply Power of a Point Theorem here. Let the radius of the larger circle be $x$ . We have $(x-3)^2=x(x-4)$ . Solving this yields $x=\dfrac{9}{2}$ . We know that $\overline{AB}$ is a diameter of $\circ O$ , so it has length $2 \cdot \dfrac{9}{2} = 9$ . We also know that $\overline{PB}$ is a diameter of the smaller circle. We have $\overline{PB}=\overline{AB}-\overline{AP} = 9-4 = 5$ , so the diameter of the smaller circle is $5$ , and the diameter is $\dfrac{5}{2} = \boxed{2.5}$

denote h=OP $(4+h)h=(1+h)^2$ $2h=1$ $h=\frac{1}{2}$ $\ r=\frac{5}{2}=\boxed{2.5}$