smc

Let $R$ and $r$ be the radius of $\odot K$ and the radius of $\odot M,$ respectively. It follows that the radius of $\odot L$ is $\frac{R}{2}.$ Suppose $P$ is the foot of the perpendicular from $M$ to $\overline{KL}.$ We construct the auxiliary lines, as shown below: In right $△KPM,$ we have $KP=r$ and $KM=R-r.$ By the Pythagorean Theorem, we get $P{M}^2=(R-r{)}^2-r^2.$ In right $△LPM,$ we have $LP=\frac{R}{2}-r$ and $LM=\frac{R}{2}+r.$ By the Pythagorean Theorem, we get $P{M}^2={\left(\frac{R}{2}+r\right)}^2-{\left(\frac{R}{2}-r\right)}^2.$ We equate the expressions for $P{M}^2,$ then simplify: $$\begin{array}{rl}(R-r{)}^2-r^2& ={\left(\frac{R}{2}+r\right)}^2-{\left(\frac{R}{2}-r\right)}^2\\ \left(R^2-2Rr+r^2\right)-r^2& =\left(\frac{R^2}{4}+Rr+r^2\right)-\left(\frac{R^2}{4}-Rr+r^2\right)\\ R^2-2Rr& =2Rr\\ R^2& =4Rr\\ R& =4r.\end{array}$$ Therefore, the ratio of the area of $\odot K$ to the area of $\odot M$ is $\frac{\pi R^2}{\pi r^2}={\left(\frac{R}{r}\right)}^2=\boxed{16}.$