2014 NJC CT Q5

\(PQ\) is the diameter of a semicircle with centre \(C\) as shown above. \(O\)is the origin and \(\overrightarrow{OP}=\mathbf{p}\) and \(\overrightarrow{OQ}=\mathbf{q}\). The tangent to the semicircle drawn from \(O\) touches the semicircle at \(T\).

1. State, in terms of \(\mathbf{p}\) and \(\mathbf{q}\), the position vector of \(C\), and hence show that the radius of the semicircle is \(\frac{1}{2}\left| \mathbf{p}-\mathbf{q} \right|\) units.

By considering the triangle \(OTC\), show that \(O{{T}^{2}}=\mathbf{p}\cdot \mathbf{q}\).

2. Give a geometrical interpretation of \(\frac{\left| \overrightarrow{OC}\cdot \overrightarrow{OT} \right|}{OT}\) and write down its value in terms of \(\mathbf{p}\) and \(\mathbf{q}\).
3. Show that \(\left| \overrightarrow{OC}\times \overrightarrow{OT} \right|=\frac{1}{m}\left| \mathbf{p}-\mathbf{q} \right|\sqrt{\mathbf{p}\cdot \mathbf{q}}\), where \(m\) is an integer to be determined.