1. Identify the given elements and their properties:
- $\triangle ABC$ is an acute triangle.
- Circle through $B$ touching $AC$ at $A$ has center $P$.
- Circle through $A$ touching $BC$ at $B$ has center $Q$.
- $R$ is the circumradius and $O$ is the circumcenter of $\triangle ABC$.

2. **Establish the collinearity of $P, Q,$ and $O$:**
- Since $P$ and $Q$ are centers of circles touching $AC$ and $BC$ respectively, and both circles pass through $A$ and $B$, $P$ and $Q$ lie on the perpendicular bisector of $AB$.
- The circumcenter $O$ of $\triangle ABC$ also lies on the perpendicular bisector of $AB$.
- Therefore, $P, Q,$ and $O$ are collinear.

3. **Calculate the areas involving $P$ and $Q$:**
- The area of $\triangle AOB$ is given by:
$$S_{AOB} = \frac{1}{2} AB \cdot OQ \cdot \sin(\angle AOB)$$
- Similarly, the area of $\triangle AOP$ is:
$$S_{AOP} = \frac{1}{2} AB \cdot OP \cdot \sin(\angle AOP)$$

4. **Relate the areas to the product $OP \cdot OQ$:**
- Using the areas calculated, we have:
$$OP \cdot OQ = \frac{S_{AOBQ} \cdot S_{AOBP}}{R^2 \cdot \sin^2 C}$$

5. Analyze the angles and radii:
- The angle $\angle PBA = \angle PAB = 90^\circ - \angle A$, thus $\angle APB = 2 \angle A$.
- Let $R_1$ be the circumradius of the circle with center $P$ in $\triangle APB$:
$$4R^2 \sin^2 C = 2R_1^2 - 2R_1^2 \cos 2A \Rightarrow R_1 = \frac{R \sin C}{\sin A}$$

6. **Calculate the areas involving $P$ and $Q$ in terms of $R$:**
- The area $S_{APBO}$ is:
$$S_{APBO} = S_{APB} + S_{AOB} = \frac{R^2 \sin 2C}{2} + \frac{R^2 \sin^2 C \sin 2A}{\sin^2 A} = \frac{R^2 \sin C \cdot \sin B}{\sin A}$$
- Similarly, the area $S_{AQBO}$ is:
$$S_{AQBO} = \frac{R^2 \sin C \cdot \sin A}{\sin B}$$

7. **Combine the areas to find the product $OP \cdot OQ$:**
- The product of the areas is:
$$S_{APBO} \cdot S_{AQBO} = R^4 \sin^2 C$$
- Therefore:
$$OP \cdot OQ = R^2$$

$\blacksquare$

The final answer is $\boxed{ R^2 = OP \cdot OQ }$