4 Mutually Tangent Circles
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Circle centered at A has radius `R_1.` Curvature is `k_1 = 1/R_1.` Circle centered at B has radius `R_2.` Curvature is `k_2 = 1/R_2.` Circle centered at C has radius `R_3.` Curvature is `k_3 = 1/R_3.` `A = (0,R_1)` `B=(0,-R_2)` Coordinates of `C = (x,y)` can be calculated as follows The area of triangle ABC can be calculated 2 ways `A= 1/2(R_1+R_2)x = sqrt((R_1+R_2+R_3)R_1R_2R_3)` `x = 2sqrt((R_1+R_2+R_3)R_1R_2R_3)/(R_1+R_2)` To calculate the y coordinate we look at the distances from A to C and from B to C. `x^2+(y-R_1)^2 = (R_1+R_3)^2` `x^2+(y+R_2)^2 = (R_2+R_3)^2` Solve simultaneously `(y+R_2)^2-(y-R_1)^2 = (R_2+R_3)^2-(R_1+R_3)^2` `y^2+2R_2y+R_2^2-y^2+2R_1y-R_1^2=R_2^2+2R_2R_3+R_3^2-R_1^2-2R_1R_3-R_3^2` `2y(R_2+R_1) = 2R_3(R_2-R_1)` `y=R_3(R_2-R_1)/(R_2+R_1)` |
Recap
`A= (0,R_1)=(0,1/k_1)`
`B=(0,-R_2)=(0,-1/k_2)`
`C= (2sqrt((R_1+R_2+R_3)R_1R_2R_3)/(R_1+R_2),R_3(R_2-R_1)/(R_2+R_1))`
Convert C from radii to curvatures
`C= (2sqrt((R_1+R_2+R_3)R_1R_2R_3)/(R_1+R_2),R_3(R_2-R_1)/(R_2+R_1))`
`= (2sqrt((1/k_1+1/k_2+1/k_3)1/k_1 1/k_2 1/k_3)/(1/k_1+1/k_2),1/k_3(1/k_2-1/k_1)/(1/k_2+1/k_1))`
`= (2sqrt((k_2k_3+k_1k_3+k_1k_2)/(k_1k_2k_3)1/(k_1k_2k_3))/((k_2+k_1)/(k_1k_2)),1/k_3((k_1-k_2)/(k_1k_2))/((k_2+k_1)/(k_1k_2)))`
`= ((2sqrt(k_2k_3+k_1k_3+k_1k_2))/(k_3(k_2+k_1)),(k_1-k_2)/(k_3(k_2+k_1)))`
Now, D is a linear combination of A and C
`D = R_3/(R_1+R_3)A+R_1/(R_1+R_3)C`
`D = (1/k_3)/(1/k_1+1/k_3)A+(1/k_1)/(1/k_1+1/k_3)C`
`D = (k_1)/(k_1+k_3)A+(k_3)/(k_1+k_3)C`
`D = (k_1)/(k_1+k_3)(0,1/k_1)+(k_3)/(k_1+k_3) ((2sqrt(k_2k_3+k_1k_3+k_1k_2))/(k_3(k_2+k_1)),(k_1-k_2)/(k_3(k_2+k_1)))`
`D = ((2sqrt(k_2k_3+k_1k_3+k_1k_2))/((k_1+k_3)(k_2+k_1)),1/(k_1+k_3)+(k_1-k_2)/((k_1+k_3)(k_2+k_1)))`
`D = ((2sqrt(k_2k_3+k_1k_3+k_1k_2))/((k_1+k_3)(k_2+k_1)),(2k_1)/((k_1+k_3)(k_2+k_1)))`
Likewise E is a linear combination of B and C
`E = R_3/(R_2+R_3)B+R_2/(R_2+R_3)C`
`E = (1/k_3)/(1/k_2+1/k_3)B+(1/k_2)/(1/k_2+1/k_3)C`
`E = (k_2)/(k_2+k_3)A+(k_3)/(k_2+k_3)C`
`E = (k_2)/(k_2+k_3)(0,-1/k_2)+(k_3)/(k_2+k_3) ((2sqrt(k_2k_3+k_1k_3+k_1k_2))/(k_3(k_2+k_1)),(k_1-k_2)/(k_3(k_2+k_1)))`
`E = ((2sqrt(k_2k_3+k_1k_3+k_1k_2))/((k_2+k_3)(k_2+k_1)),-1/(k_2+k_3)+(k_1-k_2)/((k_2+k_3)(k_2+k_1)))`
`E = ((2sqrt(k_2k_3+k_1k_3+k_1k_2))/((k_2+k_3)(k_2+k_1)),(-2k_2)/((k_2+k_3)(k_2+k_1)))`
We now use an inversion across the unit circle.
The circle centered at A goes to the line `y=1/(2R_1)=1/2k_1`
The circle centered at B goes to the line `y=-1/(2R_2)=-1/2k_2`
Since D is on the circle centered at A we know that its image D' will have a y coordinate of `1/2k_1.`
`D' = 1/2k_1*1/D_y((2sqrt(k_2k_3+k_1k_3+k_1k_2))/((k_1+k_3)(k_2+k_1)),(2k_1)/((k_1+k_3)(k_2+k_1)))`
`D' = ((sqrt(k_2k_3+k_1k_3+k_1k_2))/(2),1/2k_1)`
Likewise E is on the circle centered at B, so we know that its image E' will have a y coordinate of `-1/2k_2.`
`E' = -1/2k_2*1/E_y ((2sqrt(k_2k_3+k_1k_3+k_1k_2))/((k_2+k_3)(k_2+k_1)),(-2k_2)/((k_2+k_3)(k_2+k_1)))`
`E' = ((sqrt(k_2k_3+k_1k_3+k_1k_2))/(2),-1/2k_2)`
It is important to notice that the x-coordinates of D' and E' are the same. This makes sense when we note that the circle centered at C is being transformed into a circle that is tangent to the two lines `y=1/2k_1` and `y=-1/2k_2,` with D' as E' the points of tangency.
The center of this transformed circle is at ` ((sqrt(k_2k_3+k_1k_3+k_1k_2))/(2),1/4(k_1-k_2))`
The center of the transformed 4th circle is at ` Q' = ((sqrt(k_2k_3+k_1k_3+k_1k_2))/(2) + 1/2(k_1+k_2),1/4(k_1-k_2))` and the radius is `r=1/4(k_1+k_2).`
Let `h = abs(Q').`
The point on the transformed 4th circle that is farthest from the origin is at a distance of `h+r.` So, the closest point to the origin on circle 4 is at a distance of `1/(h+r).`
The point on the transformed 4th circle that is closest to the origin is at a distance of `h-r.` So, the farthest point from the origin on circle 4 is at a distance of `1/(h-r).`
This means that `r_4 = 1/2(1/(h-r)-1/(h+r))= ((h+r)-(h-r))/(2(h+r)(h-r)) = r/(h^2-r^2).`
`r_4 =(1/4(k_1+k_2))/( ((sqrt(k_2k_3+k_1k_3+k_1k_2))/(2) + 1/2(k_1+k_2))^2+(1/4(k_1-k_2))^2-(1/4(k_1+k_2)^2))`
`r_4 =(4(k_1+k_2))/( (2sqrt(k_2k_3+k_1k_3+k_1k_2) + 2(k_1+k_2))^2+(k_1-k_2)^2-(k_1+k_2)^2)`
`r_4 =(4(k_1+k_2))/( (2sqrt(k_2k_3+k_1k_3+k_1k_2) + 2(k_1+k_2))^2-4k_1k_2)`
`r_4 =((k_1+k_2))/( (sqrt(k_2k_3+k_1k_3+k_1k_2) + (k_1+k_2))^2-k_1k_2)`
`r_4 =((k_1+k_2))/( k_2k_3+k_1k_3+k_1k_2 +2(k_1+k_2)sqrt(k_2k_3+k_1k_3+k_1k_2) + (k_1+k_2)^2-k_1k_2)`
`r_4 =((k_1+k_2))/( k_2k_3+k_1k_3 +2(k_1+k_2)sqrt(k_2k_3+k_1k_3+k_1k_2) + (k_1+k_2)^2)`
`r_4 =((k_1+k_2))/( (k_1+k_2)(k_3 +2sqrt(k_2k_3+k_1k_3+k_1k_2) + (k_1+k_2))`
`r_4 =1/(k_1+k_2+k_3 +2sqrt(k_2k_3+k_1k_3+k_1k_2))`
Note: this formula is symmetric in `k_1,` `k_2,` and `k_3.`
It looks better as
`k_4 =k_1+k_2+k_3 +2sqrt(k_2k_3+k_1k_3+k_1k_2)`
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