$|\vec{a}|=4,|\vec{b}|=2|\vec{c}|,(\vec{a}-\vec{c})\cdot (\vec{b}-\vec{c})=3$, try minimize $|\vec{a}-\vec{b}|$

Question

$\vec{a},\vec{b},\vec{c}$ are vectors on a plane, which satisfy

$$ |\vec{a}|=4,|\vec{b}|=2|\vec{c}| $$

and the equation

$$ (\vec{a}-\vec{c})\cdot (\vec{b}-\vec{c})=3 $$

Find the minimum value of $|\vec{a}-\vec{b}|$.

I've checked the reference solution of this problem, which notes

$$ \vec a=\vec{OA},\vec{b}=\vec{OB},c=\vec{OC} $$

then notes

$$ \vec{OM}=\frac{1}{2}(\vec{OA}+\vec{OB}) $$

so we can translate the equation $(\vec{a}-\vec{c})\cdot (\vec{b}-\vec{c})=3$ into

$$ |\vec{CM}|^2-\frac{1}{4}|\vec{a}-\vec{b}|^2=3 $$

(It's better to draw a picture there to show the geometric motivation, but my technique is limited. I'm sorry...)

So $|\vec{a}-\vec{b}|$ get the minimum value iff $|\vec{CM}|$ get the minimum value.

But the next step just assumes $O,C,M$ were on the same line. It really confuses me, because this step is naive only if C was free to move on the whole circle determined by the half length of $\vec{b}$. However, if $\vec{a},\vec{b}$ was determined, the $\vec{c}$ was also determined, instead of being wherever of a circle. I've tried another order to determine the variables but it didn't work. I'm seeking for your assitance, or you can just show another solution all by yourself.

EDIT.

Answer 1

Point $C$ lies on the circle of centre $O$ and radius ${1\over2}|\vec b|$, and also on the circle of centre $M$ and radius $\sqrt{3+{1\over4}|\vec b-\vec a|^2}$. If $B$ is too close to $A$ then those circles don't intersect and $C$ doesn't exist. The limiting case is when the circles are tangent, hence to find the minimum of $|\vec b-\vec a|$ we can consider only those positions of $B$ such that the circles are tangent, their tangency point being $C$. But in this case $C$ lies between the centres of the circles, i.e. points $OCM$ are aligned.

EDIT.

If we set $AM=r$ (i.e. $AB=2r$), $\angle OAM=\theta$, we have: $$ \begin{align} OB^2 &=4r^2+16-16r\cos\theta\\ OM^2 &=r^2+16-8r\cos\theta\\ MC^2 &=r^2+3 \end{align} $$ and the condition of tangency is $$ {1\over2}OB+MC=OM. $$ This can be conveniently solved for $\cos\theta$: $$ \cos\theta=\frac{3-3 r^2\pm2 \sqrt{3 r^4+7 r^2-6}}{4 r} $$ and the minimum value of $r$ giving a real result is that which makes the square root to vanish, namely: $$ r=\sqrt{2\over3}, $$ corresponding to: $$ \cos\theta=\sqrt{3\over32}. $$