Calculate the circle that touches three other circles Given three circles on a cartesian grid (with centres and radii known), how would you calculate the centre of the circle that touches those three?
The three known circles may have any radius length, and may touch or cross each other (but not have overlapping centres), but the calculated circle must lie externally to the three known circles.
See this online graphing tool for an example of how this looks:
https://www.desmos.com/calculator/lf1q90ymrh
Note: Imagine we have the first 3 circles as given (two red ones, plus a black one). The question is: how can we mathematically deduce the formula of the fourth circle - the purple one - that just touches the first three? In this example I added the purple circle by trial-and-error, and it is only approximate.
There is one answer against this question already. It might be correct but I don't understand how to start with 3 concrete circles - like in the link above - and then work out the fourth. I think I need someone to take that linked example, use the numbers there, and explain how to perform the maths to calculate the fourth.
 A: WLOG, one of the given circles is centered at the origin.
Express that the distance between the center of the searched circle and the other centers is the difference of the respective radii.
$$\begin{cases}(r-r_0)^2=x^2+y^2\\(r-r_1)^2=(x-x_1)^2+(y-y_1)^2\\(r-r_2)^2=(x-x_2)^2+(y-y_2)^2.\end{cases} \tag1$$
Now, subtracting these equations in pairs,
$$\begin{cases}2r(r_1-r_0)+(r_1^2-r_0^2)=2xx_1+2yy_1 -x_1^2-y_1^2\\2r(r_2-r_0)+(r_2^2-r_0^2)=2xx_2 +2yy_2-x_2^2-y_2^2.\end{cases} \tag2$$
This system is linear and you can solve for $x,y$ in terms of $r$,
$$\begin{cases}x=ar+b\\y=cr+d.\end{cases} \tag3$$
Plug this in the first equation and get the quadratic
$$(r-r_0)^2=(ar+b)^2+(cr+d)^2. \tag4$$

Note that this gives you for the same "price" the construction of the circumcircle of three points, by setting $r_0=r_1=r_2=0$.
$$\begin{cases}0=2xx_1+2yy_1 -x_1^2-y_1^2\\0=2xx_2 +2yy_2-x_2^2-y_2^2.\end{cases}$$
$$r^2=b^2+d^2=x^2+y^2.$$
A: $$(x-a)^2+(y-b)^2=r^2$$
$$(x-c)^2+(y-d)^2=s^2$$
$$(x-e)^2+(y-f)^2=t^2$$
from 1st and second circle
Let P(x,y) be a general point
and circumscribing circle has radius k
C1 be the center of first circle and so on C2,C3
now when you draw any such circle circumscribing it you have a property that suggests
P to C1 distance
$$ d_1 = k - r$$
P to C2 distance 
$$d_2= k - s$$
now difference of distance is
$$d_1-d_2=s-r$$
which is a constant
and basic definition of hyperbola suggests this only.
now write equation for it
$$\sqrt{(x-a)^2+(y-b)^2}-\sqrt{(x-c)^2+(y-d)^2}=s-r$$
as in your case s>r
similarly get other equation as
$$\sqrt{(x-a)^2+(y-b)^2}-\sqrt{(x-e)^2+(y-f)^2}=t-r$$
now solve these above 2 equations to get the center of circle
now pick any of the circle say C2 of radius s, then distance between this calculated center and center of circle C2 added to the s gives the radius of the circle
you will get your answer as given below
https://www.desmos.com/calculator/w0fnc0xeqd
