A circle of radius $r$ has chords $\overline{AB}$ of length $10$ and $\overline{CD}$ of length $7$.  When $\overline{AB}$ and $\overline{CD}$ are extended through $B$ and $C$, respectively, they intersect at $P$, which is outside the circle.  If $\angle APD = 60^{\circ}$ and $BP = 8$, then $r^{2} =$

$ \textbf{(A)}\ 70\qquad\textbf{(B)}\ 71\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 73\qquad\textbf{(E)}\ 74 $
1. **Using the Power of a Point Theorem:**
   - According to the Power of a Point Theorem, for point \( P \) outside the circle, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord.
   - Given \( PB = 8 \) and \( AB = 10 \), we have \( PA = PB + AB = 8 + 10 = 18 \).
   - Let \( PC = x \) and \( PD = x + 7 \).
   - Applying the Power of a Point Theorem:
     \[
     PB \cdot PA = PC \cdot PD \implies 8 \cdot 18 = x \cdot (x + 7)
     \]
     \[
     144 = x^2 + 7x
     \]
     \[
     x^2 + 7x - 144 = 0
     \]
   - Solving this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
     \[
     x = \frac{-7 \pm \sqrt{49 + 576}}{2} = \frac{-7 \pm 25}{2}
     \]
     \[
     x = 9 \quad \text{(since length cannot be negative)}
     \]

2. **Using the Law of Cosines in \(\triangle PAC\):**
   - We know \( PA = 18 \), \( PC = 9 \), and \(\angle APD = 60^\circ\).
   - Applying the Law of Cosines in \(\triangle PAC\):
     \[
     AC^2 = PA^2 + PC^2 - 2 \cdot PA \cdot PC \cdot \cos(60^\circ)
     \]
     \[
     AC^2 = 18^2 + 9^2 - 2 \cdot 18 \cdot 9 \cdot \frac{1}{2}
     \]
     \[
     AC^2 = 324 + 81 - 162
     \]
     \[
     AC^2 = 243
     \]
     \[
     AC = \sqrt{243} = 9\sqrt{3}
     \]

3. **Using the Law of Cosines in \(\triangle PAD\):**
   - We know \( PA = 18 \), \( PD = 16 \), and \(\angle APD = 60^\circ\).
   - Applying the Law of Cosines in \(\triangle PAD\):
     \[
     AD^2 = PA^2 + PD^2 - 2 \cdot PA \cdot PD \cdot \cos(60^\circ)
     \]
     \[
     AD^2 = 18^2 + 16^2 - 2 \cdot 18 \cdot 16 \cdot \frac{1}{2}
     \]
     \[
     AD^2 = 324 + 256 - 288
     \]
     \[
     AD^2 = 292
     \]
     \[
     AD = \sqrt{292}
     \]

4. **Finding the radius \( r \):**
   - Since \( \triangle PAC \) is a 30-60-90 triangle, we know that \( \angle PCA = 90^\circ \).
   - The radius \( r \) of the circle is half the length of the diameter \( AD \).
   - Therefore, \( r = \frac{AD}{2} = \frac{\sqrt{292}}{2} = \sqrt{73} \).
   - Hence, \( r^2 = 73 \).

The final answer is \( \boxed{73} \).