Sina drew square \(ABCD\) with side length \(6\) cm on a piece of paper and passed the paper to Theo. Theo drew a circle on top of the square so that the circle passes through \(A\) and \(D\), and the circle is tangent to side \(BC\) at point \(P\).
Determine the radius of the circle.
Note: You may find the following known result about circles useful:
If a line is tangent to a circle, then the perpendicular to that line at the point of tangency passes through the centre of the circle.
Let \(O\) be the centre of the circle and \(r\) be the radius. Construct line segment \(PQ\) perpendicular to \(CB\) with \(Q\) on side \(AD\) of the square. Since \(CB\) is tangent to the circle with point of tangency \(P\), \(PQ\) must pass through the centre of the circle, \(O\). Therefore, \(PO=r\).
Since \(PQ\perp BC\), \(PQ\parallel AB,\) and \(PQ=AB=6\), then \(QO=PQ-PO=6-r\). Since \(A\) and \(D\) are on the circle, \(AO=DO=r\).
Using the Pythagorean Theorem, \(AQ^2=AO^2-QO^2=r^2-(6-r)^2\) and \(DQ^2=DO^2-QO^2=r^2-(6-r)^2\). Therefore, \(AQ^2=DQ^2\) and \(AQ=DQ\) follows. Since \(AQ=DQ\) and \(AQ+QD=AD=6\), we can substitute to obtain \(AQ+AQ=2AQ=6\) or \(AQ=3\).
Using the Pythagorean Theorem in \(\triangle AQO\), \[\begin{aligned} AO^2&=AQ^2+QO^2\\ r^2&=3^2+(6-r)^2\\ r^2&=9+36-12r+r^2\\ 12r&=45\\ r&=\frac{45}{12}\\ r&=3.75 \end{aligned}\] Therefore, the radius of the circle is \(3.75\) cm.