

last edited 4 years ago by test1 
1  
Editor: test1
Time: 2020/02/10 18:11:11 GMT+0 

Note: 
changed:  This is motivated by discussion in Sage mailing list. We are given sum of ellipsis and a strip symmetric with respect to one of the axis of the ellipsis. We want to find circle with center on the line x = D which is tangent both to ellipsis and the strip. More precisely, we want to find common point (x, r) and radius R of the circle. We get equations (for convenience written as polynomials): \begin{axiom} )set output algebra on )set output tex off eq1 := (x/r1)^2 + (y/r2)^2  1 eq2 := (xD)^2+(yd/2R)^2  R^2 eq3 := (xD)/(yd/2R)  (x/y)*(r2/r1)^2 \end{axiom} The first is equation of ellipsis, the second of the circle, third is tangency condition. This system can be solved, but it would take long time, the answer is large and would be hard to read. So first we replace r, r2, D, d by sensible concrete values: \begin{axiom} eq1e := eval(eq1, [r1 = 11/10, r2 = 7/10, d = 15/100, D = 2]) eq2e := eval(eq2, [r1 = 11/10, r2 = 7/10, d = 15/100, D = 2]) eq3e := eval(eq3, [r1 = 11/10, r2 = 7/10, d = 15/100, D = 2]) \end{axiom} Now we can solve \begin{axiom} rese := solve([eq1e, eq2e, eq3e], [x, y, R]) \end{axiom} Not small, but we see that x and y are expressed in terms of R and for R we get equation of degree 6. Let us see numeric solutions for R: \begin{axiom} solve(rese(1)(3), 1.0e15) \end{axiom} So there are two real solutions, positive one solve our problem, negative represents circle such that has interior (a disc) has intersection with interior of the strip. This looks unnatural, but is correct solution of the equations we gave. To exclude it we would have to add an inequality, say y > 0.
This is motivated by discussion in Sage mailing list. We are given sum of ellipsis and a strip symmetric with respect to one of the axis of the ellipsis. We want to find circle with center on the line x = D which is tangent both to ellipsis and the strip. More precisely, we want to find common point (x, r) and radius R of the circle. We get equations (for convenience written as polynomials):
(1) > )set output algebra on
)set output tex off
eq1 := (x/r1)^2 + (y/r2)^2  1
2 2 2 2 2 2 r1 y + r2 x  r1 r2 (1)  2 2 r1 r2
eq2 := (xD)^2+(yd/2R)^2  R^2
2 2 1 2 2 (2) y + ( d  2 R)y + x  2 D x +  d + R d + D 4
eq3 := (xD)/(yd/2R)  (x/y)*(r2/r1)^2
2 2 2 1 2 (( r2 + r1 )x  D r1 )y + ( d + R)r2 x 2 (3)  2 2 1 2 r1 y + (  d  R)r1 y 2
The first is equation of ellipsis, the second of the circle, third is tangency condition.
This system can be solved, but it would take long time, the answer is large and would be hard to read. So first we replace r, r2, D, d by sensible concrete values:
eq1e := eval(eq1,[r1 = 11/10, r2 = 7/10, d = 15/100, D = 2])
100 2 100 2 (4)  y +  x  1 49 121
eq2e := eval(eq2,[r1 = 11/10, r2 = 7/10, d = 15/100, D = 2])
2 3 2 3 6409 (5) y + ( 2 R  )y + x  4 x +  R +  20 20 1600
eq3e := eval(eq3,[r1 = 11/10, r2 = 7/10, d = 15/100, D = 2])
72 49 147 ( x  2)y + ( R + )x 121 121 4840 (6)  2 3 y + ( R  )y 40
Now we can solve
rese := solve([eq1e,eq2e, eq3e], [x, y, R])
(7) [ [ x = 21647170639872527456000 5 563398654925279868893800 4  R   R 11314902455332490895259929 33944707365997472685779787 + 227978377910623361179980 3 543170461396404207065460341 2  R   R 3771634151777496965086643 1086230635711919125944953184 + 103497888021194504792764569 18719246207620127101109311295  R +  2413845857137598057655451520 7724306742840313784497444864 ,
y = 8578201025986535925690800000 5   R 481302005742478165211671599873 + 687645572606252172049144000 4   R 160434001914159388403890533291 + 569404955344370090785647085975 3   R 962604011484956330423343199746 + 231817205837663178957237643885 2  R 855648010208850071487416177552 + 15047964589394026276064984240695 56533764389483354647866157177923  R +  6845184081670800571899329420416 273807363266832022875973176816640 ,
6 11937 5 16384102559 4 77692853121 3 1498249076221 2 R +  R +  R   R   R 60760 476358400 4763584000 15243468800 + 66093598324107 2139938711851329   R   3810867200000 48779100160000 = 0 ] ]
Not small, but we see that x and y are expressed in terms of R and for R we get equation of degree 6. Let us see numeric solutions for R:
solve(rese(1)(3),1.0e15)
(8) [R =  1.4998115943_024088459,R = 1.9728606253_044955743]
So there are two real solutions, positive one solve our problem, negative represents circle such that has interior (a disc) has intersection with interior of the strip. This looks unnatural, but is correct solution of the equations we gave. To exclude it we would have to add an inequality, say y > 0.