AO is a line segment containing the radii of both circles, which are AK and OE, so AO = AK + OE = 9.
Extend AO to a point P on the line KE. Then we get two similar triangles APK and OPE. Let [tex]x[/tex] be the length of KE, [tex]y[/tex] the length of EP, and [tex]z[/tex] the length of OP.
By similarity, we have
[tex]\dfrac{OE}{AK}=\dfrac{EP}{KP}=\dfrac{OP}{AP}\iff\dfrac45=\dfrac y{y+x}=\dfrac z{z+9}[/tex]
OPE is a right triangle, so
[tex]OP^2=OE^2+EP^2\implies z=\sqrt{16+y^2}[/tex]
Now,
[tex]\dfrac45=\dfrac y{y+x}\implies 4(y+x)=5y\implies 4x=y[/tex]
and we also have
[tex]z=\sqrt{16+(4x)^2}=\sqrt{16+16x^2}=4\sqrt{1+x^2}[/tex]
Substituting this into the expression containing [tex]z[/tex] gives us an equation that we can solve for [tex]x[/tex]:
[tex]\dfrac45=\dfrac z{z+9}=\dfrac{4\sqrt{1+x^2}}{4\sqrt{1+x^2}+9}[/tex]
[tex]16\sqrt{1+x^2}+36=20\sqrt{1+x^2}[/tex]
[tex]9=\sqrt{1+x^2}[/tex]
[tex]\implies KE=x=\sqrt{80}=3\sqrt{10}[/tex]
