First, we find let $AB=c,BC=a,AC=b,$ and $\frac{a+b+c}{2}=s.$ This means that $s-c=8,s-b=10.$ Adding, we find that $a=18$ and subtracting, we find that $c-b=2.$ Now, we find that $[ABC]=4s.$ From Heron's formula, we have $$\sqrt{s(s-a)(s-b)(s-c)}=4s\Leftrightarrow\sqrt{5(s-a)}=\sqrt{s}.$$Squaring, we have $5(s-a)=s.$ In other words, $a=\frac{4s}{5}.$ Plugging in $a=18,$ we find that $s=\frac{45}{2}.$ Now, we have $b+c=45-18=27.$ Now, combining this with $c-b=2,$ we conclude that $b=AC=\boxed{\frac{25}{2}}$ and $c=AB=\boxed{\frac{29}{2}}.$

Let the circle touch $AB$ at $E$ and let the circle touch $AC$ at $F$. Set $AE=AF=x$. Then $\frac{\sqrt{(x+18) \cdot 8 \cdot 10 \cdot x}}{x+18}=4$, or $80x(x+18)=16(x+18)^2 \implies 5x=x+18 \implies x = \frac{9}{2}$. We hence have $AB=\boxed{\frac{29}{2}}, AC=\boxed{\frac{25}{2}}$.