Distance to a Circle

From Power of a Point , we know the following theorem is true:

"The theorem of intersecting secants (or secant-secant power theorem) states that if PQ and RS are chords of a circle which intersect at a point A outside the circle, then

$AP \times AQ = AR \times AS$ ."

Thus, in this problem, we know that $PA \times PB = PC \times PD$ , so $PD= \frac{PA \times PB}{PC} = \frac{44 \times 294}{56} = \boxed{231}$ , which is the desired number.

P A · P B = P C · P D

Note that $A$ is between $B$ and $P$ . Since $\angle CAB$ and $\angle CDB$ are opposite angles of a cyclic quadrilateral, their sum is $180^\circ$ . Hence, $\angle PAC = 180^\circ - \angle CAB = \angle BDC = \angle BDP$ . Likewise, $\angle PCA = 180^\circ - \angle ACD = \angle PBD$ . As such, by angle-angle-angle, triangles $PAC$ and $PDB$ are similar. Thus $\frac {PA}{PD} = \frac {PC}{PB}$ $\Rightarrow PD = \frac {PA \cdot PB} {PC } = \frac {44 \cdot 294} {56} = 231$ .