Consider the right triangle OAT where O is the origin and T is the point where the circle is tangent to the graph of \(y = k\left| x \right|\) such that \(OA = a\) and \(AT = b\). Because the slope of the right side of the graph of \(y = k\left| x \right|\) is \(k\), then \(\frac{a}{b} = \frac{k}{1}\). Now consider right triangle CTO. Triangles OAT and CTO are similar (corresponding sides proportional); so, \(\frac{a}{b} = \frac{k}{1} = \frac{{OT}}{{CT}}\). Since CT is a radius of the unit circle then \(\frac{k}{1} = \frac{{OT}}{1}\;\;\; \Rightarrow \;\;\;OT = k\).