Determine $z\in\mathbb{C}$, $r>0$ so that $0,1,2+i\in \partial B_r(z)$

I'm struggling to see for a method to start this question.
It looks like a question related with mobius transforms.
We have studied about determining the mobius transform when points from the domain are given. For example, getting the cross ratio.

But here I don't see any points from the domain and also a method to determine the center when the mobius transform is given.

Help would be appreciated

• Hint: $r$ can be anything greater than half the distance from $0$ to $2+i$. – Robert Israel Apr 15 at 0:23
• Can you please explain how should I proceed afterwards – Charith Jeewantha Apr 15 at 1:38
• Given $r$, $z$ could be anything in the intersection of the disks of radius $r$ around $0$, $1$ and $2+i$. – Robert Israel Apr 15 at 2:28
• On the other hand, you could first choose any $z$, and then let $r$ be greater than the maximum of $|z|$, $|z-1|$ and $|z-(2+i)|$. – Robert Israel Apr 15 at 2:35
• OK, so you want $z$ to be equidistant from $0$, $1$ and $2+i$. Construct the perpendicular bisectors of the line segments $[0,1]$ and $[1, 2+i]$, and see where they intersect. – Robert Israel Apr 15 at 4:07

With $$z = x +iy$$ you have the three equations
1. $$\lvert z - 0 \rvert = r$$. i.e. $$x^2 + y^2 = r^2$$
2. $$\lvert z - 1 \rvert = r$$, i.e. $$(x - 1)^2 + y^2 = x^2 - 2x + 1 + y^2 = r^2$$
3. $$\lvert z - (2+i) \rvert = r$$, i.e. $$(x - 2)^2 + (y - 1)^2 = (x - 2)^2 + y^2 - 2y + 1 = r^2$$
Subtracting 2.from 1. you get $$2x - 1 = 0$$, hence $$x = \frac{1}{2}$$. Thus $$y^2 = r^2 - \frac{1}{4}$$. Inserting this in 3. yields $$\frac{9}{4} + r^2 - \frac{1}{4} - 2y +1 = r^2$$ which means $$y = \frac{3}{2}$$.
Then you get $$\frac{1}{4} + \frac{9}{4} = r^2$$, i.e. $$r = \sqrt{\frac{5}{2}}$$.