By "touch it internally", does it mean that they are internally tangent?

Overkill method: Let X be the intersection of $c_1$ and $c_2$. Consider the inversion of $c$, since $\angle OAX' = 90$, $\angle OBX' = 90$, so $(OAX'B)$ and thus AXB collinear.

Proper method: Since AO and BO are the diameters of $c_1$ and $c_2$ (If the center is Y then AYO collinear by tangency, so O is the antipode of A, same for the other circle) $\angle AXO = 90$, $\angle BXO = 90$, so $\angle AXB = 180$ and we are done.

Coord bash overkill method (actually pretty cool): Describe $c$ by $x^2+y^2=4$ and fix $c_1$ at $(x-1)^2+y^2=1$. This means that $A=(2,0)$. Then let $c_2$ be $(x-\cos{\theta})^2+(y-\sin{\theta})^2=1$, and $B=(2\cos{\theta},2\sin{\theta})$. Thus $\overline{AB}$ has slope $\frac{\sin{\theta}}{\cos{\theta}-1}$ and its graph can be described by. The intersection points between $c_1$ and $c_2$ are $(0,0)$ (since they must intersect at the center), and its graph is $y=\frac{\sin{\theta}}{\cos{\theta}-1}(x-2)$ The intersection points between $c_1$ and $c_2$ are $(0,0)$ and another point, which is the intersection between $y=(\frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}x)=(\tan{\frac{\theta}{2}})x$ and $c_1$. This yields the equation $(x-1)^2+y^2=1 \Rightarrow (x-1)^2+(\tan{\frac{\theta}{2}}*x)^2=1 \Rightarrow x^2-2x+\tan^2{\frac{\theta}{2}}*x^2=0$. Since $x \neq 0$ (we accounted for that already) we can divide and get $x(1+\tan^2{\frac{\theta}{2}})=2 \Rightarrow x=\frac{2}{1+\tan^2{\frac{\theta}{2}}}$ and $y=\frac{2\tan{\frac{\theta}{2}}}{1+\tan^2{\frac{\theta}{2}}}$. We now must show that $y=\frac{\sin{\theta}}{\cos{\theta}-1}(x-2)$ passes through $(\frac{2}{1+\tan^2{\frac{\theta}{2}}},\frac{2\tan{\frac{\theta}{2}}}{1+\tan^2{\frac{\theta}{2}}})$. Note that $\frac{2}{1+\tan^2{\frac{\theta}{2}}}=\frac{2\cos^2{\frac{\theta}{2}}}{\cos^2{\frac{\theta}{2}}+\sin^2{\frac{\theta}{2}}}=2\cos^2{\frac{\theta}{2}}=\cos{\theta}+1$, which means that $\frac{2\tan{\frac{\theta}{2}}}{1+\tan^2{\frac{\theta}{2}}}=(\tan{\frac{\theta}{2}})(\cos{\theta}+1)=\sin{\theta}$. So the LHS of the equation is simply $\sin{\theta}$. Further the RHS is $\frac{\sin{\theta}}{\cos{\theta}-1}(\cos{\theta}-1)=\sin{\theta}$. This means that we have $\sin{\theta}=\sin{\theta}$, which is obviously true.