Let $|\vec a|=|\vec b|=2$ and $|\vec c|=1$.Find the difference between maximum and minimum possible values of $|\vec a+\vec b|$

Question

Let $|\vec a|=|\vec b|=2$ and $|\vec c|=1$. Also $(\vec a-\vec c)\cdot(\vec b-\vec c)=0$.

Find the difference between maximum and minimum possible values of $|\vec a+\vec b|$

My Attempt

$|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b=4+4+2\vec c\cdot(\vec a+\vec b)-2=6+2\vec c\cdot(\vec a+\vec b)$.

After this I am not able to proceed

But after some time I did manage to come up with a possible solution but am not able to interpret it geometrically.

$|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b=8+8\cos\theta=16\cos^2\frac{\theta}{2}$

$\Rightarrow|\vec a+\vec b|=4|\cos\frac{\theta}{2}|$

Also it is given that

$(\vec a-\vec c)\cdot(\vec b-\vec c)=0$

$\Rightarrow \vec a\cdot\vec b-\vec c\cdot(\vec a+\vec b)+|\vec c|^2=0$

$\Rightarrow 4\cos\theta-4|\cos\frac{\theta}{2}|\cos\alpha+1=0$

$\Rightarrow4(2\cos^2\frac{\theta}{2}-1)-(4\cos\alpha)|\cos\frac{\theta}{2}|+1=0$

$\Rightarrow 8\cos^2\frac{\theta}{2}-(4\cos\alpha)|\cos\frac{\theta}{2}|-3=0$

$\Rightarrow 4|\cos\frac{\theta}{2}|=\cos\alpha\pm\sqrt{\cos^2\alpha+6}$

$\Rightarrow |\vec a+\vec b|=\cos\alpha+\sqrt{\cos^2\alpha+6}$

Let $f(x)=x+\sqrt{x^2+6}$ where $x\in [-1,1]$

$f'(x)=1+\frac{x}{\sqrt{x^2+6}}=\frac{x+\sqrt{x^2+6}}{\sqrt{x^2+6}}>0$

$f(x)$ is strictly increasing.

$f_{min}=\sqrt{7}-1$ and $f_{max}=\sqrt{7}+1$

$\Rightarrow \sqrt{7}-1\leq |\vec a+\vec b|\leq \sqrt{7}+1$

I am not able to justify it geometrically till now.