Find the value of CM on the circle below

Question

In the figure, if $C$, $A$ and $N$ are points of tangency , determine $CM$. ($S:CM=2$)

I try

CNOI is a square $\therefore CN = l\boxed{}\sqrt2 \implies l_\boxed{}=3\sqrt2\\ FN.MN = DN.IN \implies 9.(6+CM) = DN.3\sqrt2\\ \therefore 18+3CM = DN\sqrt2$

out-of-scale figure

Answer 1

As can be see in figure KN is the diameter of a circle which crosses N', where N' is the projection of N on DB. So this circle is tangent on segment FM at N, that is $KN\bot FM$ , so N must be the mid point of chord FM. In this way we have:

$NM=FN=6\Rightarrow CM=6-4=2$

Answer 2

$ CN = l\boxed{}\sqrt2 \implies l_\boxed{}=3\sqrt2\\ FN.MN = DN.IN \implies 9.(6+CM) = DN.3\sqrt2\\ \therefore 18+3CM = DN\sqrt2(I)\\ \triangle OFP: OF^2 = 9^2+(3\sqrt2)^2+2.9.3\sqrt2(cos135^o)\implies OF = 3\sqrt{17}\\ \triangle DNO \sim \triangle FPO: \frac{DN}{12}=\frac{3\sqrt2}{3} \implies DN = 12 \sqrt2\\ From(I):18+3CM =12\sqrt2.\sqrt2 \implies 3CM = 6 \therefore \boxed{CM = 2} $

Answer 3

I have a solution, but it's tedious lol. I thought it would be helpful to share nonetheless.

Let $IB = a, ID = b$. The formula for the incircle radius of $\Delta IDB$ is $$ ON = \frac{2 \cdot Area(IDB)}{Perimeter(IDB)} \\ \Rightarrow 3\sqrt{2} = \frac{ab}{a^2+b^2 + \sqrt{a^2 + b^2}} \text{ } (1) $$ Also, we have: $$ 18+3CM = DN\sqrt2 \\ \Rightarrow CM = \frac{(ID - 3\sqrt2)\sqrt2 - 18}{3} \\ \Rightarrow CM = \frac{\sqrt2}{3} b - 8 \text{ } (2) $$

Using intersecting chords again, but on $FM$ and $IB$, we have:

$$ FC \cdot CM = IC \cdot CB \\ \Rightarrow 15 CM = 3\sqrt2 (IB - IC) \\ \Rightarrow 15 CM = 3\sqrt2 (a - 3\sqrt2) \\ \Rightarrow CM = \frac{\sqrt2}{5}a - \frac{6}{5} (3) $$

From $(2)$ and $(3)$, we have: $$ CM = \frac{\sqrt2}{3} b - 8 = \frac{\sqrt2}{5}a - \frac{6}{5} (4) $$

$(1)$ and $(4)$ gives the following system of equation: $$ \begin{cases} 3\sqrt{2} = \frac{ab}{a^2+b^2 + \sqrt{a^2 + b^2}} \\ \frac{\sqrt2}{3} b - 8 = \frac{\sqrt2}{5}a - \frac{6}{5} \end{cases} $$

After plugging in $b = \frac{3}{5} a + \frac{51 \sqrt2}{5}$ (which can be derived from $(4)$) into $(1)$, we have a quartic equation. I ran each step through Wolframalpha (multiply, square both sides, ...) and arrived at this equation: $$ \frac{-a^4}{50} - \frac{9 \sqrt2 a^3}{25} + \frac{11 a^2}{5} + \frac{1224 \sqrt2 a}{25} = 0 $$ All of its solutions are $a \in \{ -17 \sqrt2, -9 \sqrt2, 0, 8 \sqrt2 \}$. Therefore $a = IB = 8 \sqrt2$, and CM can be calculated using $(3)$.