# P.o.t.W. #8 Solution

SOLUTIONConsider 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...

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