How to draw $4$ touching circles I am trying to figure out a way for drawing $4$ touching circles like this:

Would appreciate help of any kind.
 A: If you don't have any conditions other than four touching circles, I could imagine constructing it in GeoGebra (at least the way I imagine it working).
Construct circles $O_1$ and $O_3$ with radii $r_1$ and $r_3$.  Now, the locus of centers of circles that would be tangent to both of them would be points that are $r_1-r_3$ units closer to $O_3$ than $O_1$.  That locus is a hyperbola with foci at $O_1$ and $O_3$.  Then choose any $O_2$ and $O_4$ on that curve and finish it up.

So, you can do this in GeoGebra without too much effort.  Follow these steps:


*

*Create two non-intersecting circles.  Make one bigger than the other to start

*Draw a line segment connecting the centers of the circles and draw points where that line intersects the two circles.

*Construct the midpoint of the two circle intersection points

*Construct a hyperbola.  They ask you to click on two foci and then a point of the hyperbola.  The foci are the two circle centers and the point on the hyperbola will be midpoint you constructed in the last step

*Choose a point on the hyperbola.  Make sure you are using the same branch as the one that goes through the midpoint of the line segment.  (If the two original circles have different radii, it should be easy to make out.)


*

*Draw a line segment from that point to either of the two original circle centers

*Construct a circle whose center is the new point and lies on the intersection of the new line segment with the original circle.

*Repeat the preceding three steps to make a fourth circle that doesn't intersect the third one

*Hide all of your auxiliary elements and enjoy your four semi-kissing circles!



If you do this well enough, you can probably even drag the centers around to ajust the entire structure in real-time.  I'll let you play with it.
A: If circles $k1 , k2 $ are fixed then positions of $k3, k4$ can be  varied. There is no unique relative position that can be determined like in 3 circles  "kiss precise" case.
This is because in the quadrilateral $O_1O_2O_3O_4$only four lengths are given when actually you need five, making it to move like a mechanism.
So if two adjacent circles are fixed then the remaining two would be wheeling up and down...
A: You can actually do this starting with $A$, $B$, $C$, and $D$, and you still have a degree of freedom... but there's a restriction on the locations of these four points: they must form a cyclic quadrilateral.  (I'm not sure why, but it works out that way - perhaps because the perpendicular sectors all meet or something).
Once we have the points in place, it's easy: make segments $a = AB$, $b = BC$, $c = CD$, $d = DA$, and their perpendicular bisectors $a'$ etc.  place a point $E$ anywhere along $a'$ (and I do mean anywhere: if it's too close it will make other circles flip sides but tangency still holds!), draw a line $EB$ and get its intersection $F$ on $b'$, $FC$ intersects $c'$ at $G$, $GD$ intersects $d'$ at $H$.  Then $E$, $F$, $G$, $H$ are the centers of four circles that meet at $A$, $B$, $C$, and $D$.
A: Well, drawing the first three circles is easy : pick a point $O_1$ and a radius $k_1$, trace the first circle $C_1$, pick a point $O_2$ and pick radius $k_2$ to be the distance from $O_2$ to $O_1$ minus $k_1$ and trace $C_2$, pick a point $O_3$ and pick radius $k_3$ to be the distance from $O_3$ to $O_2$ minus $k_2$ and trace $C_3$.
The part that’s a litte more difficult is to find a $O_4$ and $k_4$ that “fits in” the first and third circles.  For this one, pick a radius $k_4$ first.  Be reasonable, too small, and the circle cannot possibly touch both the first and the third circles, too large and you won’t be able to trace the circle inside your sheet.
Now, trace a circle $\bar C_1$ centered in $O_1$ and of radius $k_1 + k_4$ and another circle circle $\bar C_3$ centered in $O_3$ and of radius $k_3 + k_4$.
Where  $\bar C_1$ and  $\bar C_3$ meets, it is a point at distance $k_1 + k_4$ of $O_1$ and $k_3 + k_4$ of $O_3$.  Therefore name this intersection point $O_4$ and trace the circle $C_4$ of radius $k_4$ [There should be two such intersections.  If $k_4$ is about $k_2$, one such intersection should be near $O_2$, so pick the other one to have something more like your picture above].  Because $O_4$ has distance $k_1 + k_4$ from $O_1$, circle $C_4$ of radius $k_4$ will touch circle $C_1$ of radius $k_1$ at one point. Likewise $C_4$ will touch $C_3$ at one point.
