Given:
 Line \[QA\left\| {PC} \right.\]
\[\begin{gathered}
  QA = 7.2cm \\
  PC = 5cm \\
\end{gathered} \]

To find the radius of the circle.

Hint:
first of all, we’ll draw a diagram and perform needful construction. then we’ll use tests of similarity of triangles to show that two triangles are similar. Also, we will take help of theorem of the geometric mean.

Complete step by step solution:
First, we need to do some construction.
Join OA and OB.

now we can observe four triangles in a given circle.
Consider \[\vartriangle AOQ\] and \[\vartriangle AOB\]
\[AO \cong AO\]→ common side
\[QA \cong BA\]→ tangents to circle from external point
\[OQ \cong OB\]→ radius of same circle
Hence \[\vartriangle AOQ \cong \vartriangle AOB\]→ by SSS test.
Consider, \[\angle AOQ \cong \angle AOB\]→ C.A.C.T.
\[\angle AOQ \cong \angle AOB\]=ά
Similarly, \[\vartriangle COP \cong \vartriangle COB\]→ by SSS test.
Consider, \[\angle COP \cong \angle COB\]→ C.A.C.T.
\[\angle COP \cong \angle COB\]=β
But, \[\angle AOQ + \angle AOB + \angle COP + \angle COB = {180^ \circ }\]
\[
  2\alpha + 2\beta = {180^ \circ } \\
  \alpha + \beta = {90^ \circ } \\
\]
Thus ,
\[
  \angle AOB + \angle COB = {90^ \circ } \\
  \angle AOC = {90^ \circ } \\
  OB \bot AC \\
\]
Using theorem of geometric mean,
\[
  O{B^2} = AB \times BC \\
   \Rightarrow O{B^2} = 7.2 \times 5 \\
   \Rightarrow O{B^2} = 36 \\
   \Rightarrow OB = 6 \\
\]

Thus the radius of the circle is 6cm.

Additional information:
1) Tangent is a line drawn from an external point of a circle such that it touches the circle in one and only one point.
2) Tangents drawn from an external point of a circle to a circle are congruent.

Note:
1) When we face such geometrical problems we need to add it with some construction so as to find the answer or to prove if anything asked.
2) SSS test is a test for determining the congruency of triangles.