Question

Given: Circles k1(A) and k2(O)ext. tangent

Answer 1

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 $x$ be the length of KE, $y$ the length of EP, and $z$ the length of OP.

By similarity, we have

$\dfrac{OE}{AK}=\dfrac{EP}{KP}=\dfrac{OP}{AP}\iff\dfrac45=\dfrac y{y+x}=\dfrac z{z+9}$

OPE is a right triangle, so

$OP^2=OE^2+EP^2\implies z=\sqrt{16+y^2}$

Now,

$\dfrac45=\dfrac y{y+x}\implies 4(y+x)=5y\implies 4x=y$

and we also have

$z=\sqrt{16+(4x)^2}=\sqrt{16+16x^2}=4\sqrt{1+x^2}$

Substituting this into the expression containing $z$ gives us an equation that we can solve for $x$:

$\dfrac45=\dfrac z{z+9}=\dfrac{4\sqrt{1+x^2}}{4\sqrt{1+x^2}+9}$

$16\sqrt{1+x^2}+36=20\sqrt{1+x^2}$

$9=\sqrt{1+x^2}$

$\implies KE=x=\sqrt{80}=3\sqrt{10}$