Problem 1638

1. Label the circles and points:
- Let $C_1$ be the circle with center $O$ and radius $R$.
- Let $AB$ be a chord of $C_1$ with midpoint $M$.
- Let $C_2$ be the circle with diameter $OM$ and center $U$.
- Let $T$ be a point on $C_2$.
- The tangent to $C_2$ at $T$ intersects $C_1$ at points $P$ and another point (not labeled).

2. Apply Apollonius' theorem:
- Apollonius' theorem states that for any triangle $PAB$ with midpoint $M$ of $AB$, we have:
$$PA^2 + PB^2 = 2(PM^2 + MA^2)$$

3. **Express $MA$ in terms of $R$ and $r$:**
- Since $M$ is the midpoint of $AB$, $MA = MB$.
- In $\triangle OAM$, by the Pythagorean theorem:
$$OM^2 = OA^2 - AM^2 \implies (2r)^2 = R^2 - AM^2 \implies 4r^2 = R^2 - AM^2 \implies AM^2 = R^2 - 4r^2$$

4. **Relate $PM$ and $PU$:**
- Since $U$ is the center of $C_2$ and $OM$ is the diameter, $U$ is the midpoint of $OM$.
- Therefore, $PU = PM$ because $P$ lies on the tangent to $C_2$ at $T$ and $T$ is on $C_2$.

5. **Substitute $MA^2$ and $PM$ into Apollonius' theorem:**
- Using $MA^2 = R^2 - 4r^2$:
$$PA^2 + PB^2 = 2(PM^2 + MA^2) = 2(PM^2 + R^2 - 4r^2)$$

6. **Express $PM$ in terms of $PT$:**
- Since $P$ is on the tangent to $C_2$ at $T$, the distance from $P$ to $T$ is the same as the distance from $P$ to $U$ (the radius of the tangent circle):
$$PM = PT$$

7. Combine the equations:
- Substitute $PM = PT$ into the equation:
$$PA^2 + PB^2 = 2(PT^2 + R^2 - 4r^2)$$

8. Simplify the expression:
- Distribute the 2:
$$PA^2 + PB^2 = 2PT^2 + 2R^2 - 8r^2$$

9. **Relate $R$ and $r$:**
- Since $OM$ is the diameter of $C_2$, $OM = 2r$ and $R$ is the radius of $C_1$:
$$PA^2 + PB^2 = 2PT^2 + 2R^2 - 8r^2$$

10. Final simplification:
- Notice that $2R^2 - 8r^2$ can be rewritten as $2(R^2 - 4r^2)$:
$$PA^2 + PB^2 = 2PT^2 + 2(R^2 - 4r^2)$$
- Since $R^2 - 4r^2 = MA^2$:
$$PA^2 + PB^2 = 2PT^2 + 2MA^2$$
- But $MA^2 = R^2 - 4r^2$:
$$PA^2 + PB^2 = 2PT^2 + 2(R^2 - 4r^2) = 2PT^2 + 2MA^2$$
- Finally, we get:
$$PA^2 + PB^2 = 4PT^2$$

The final answer is $\boxed{PA^2 + PB^2 = 4PT^2}$.