Tangent to a Circle

Let $r$ be the radius of $\Gamma$ . Since $PA$ is tangent to $\Gamma$ , we have $\angle PAO = 90^\circ$ (Thales' theorem). Then the Pythagorean theorem implies $\begin{aligned} PA^2 + AO^2 &= PO^2 \\ PA^2 + AO^2 &= (PD + DO)^2 \\ 42 ^2 + r^2 &= (14 + r)^2 \\ 42 ^2 + r^2 &= 196 + 28r + r^2 \\ r &= \frac{ 42 ^2 - 196} {28} \\ &= \frac{1568}{28} \\ &= 56 \\ \end{aligned}$

The line $PO$ intersects $\Gamma$ at $D$ and at other point, let's call it $E$ . Due to power of a point relative to a circumference theorem $PE \times PD = PE \times 14 = PA^2 = 42^2 = 1764 \Rightarrow PE = \frac{1764}{14} = 126$ . Now, let $r$ be the radius of the circumference $\Gamma$ then $2r = PE - PD = 126 - 14 = 112 \Rightarrow r = \frac{112}{2} = 56$