Ja: dit probleem is volledig op te lossen met herhaald toepassen van de stelling van Pythagoras.

Mooi...
ik neem aan dat je onderstaande werkwijze hebt toegepast?
[attachment=1]cirkel.png[/attachment][attachment=0]uitwerking.png[/attachment]

 r[1]  = 0.107206358681917884414824
 r[2]  = 0.063748687512154396116496
 r[3]  = 0.032821511636756146746755
 r[4]  = 0.015670122563985885102203
 r[5]  = 0.007214141551854388718512
 r[6]  = 0.003265777217972012640372
 r[7]  = 0.001467139761780204657812
 r[8]  = 0.000656847751170706173659
 r[9]  = 0.000293622807019700485942
 r[10] = 0.000131164440584095479382
 r[11] = 0.000058574551480610704541
 r[12] = 0.000026154247424269826999
 r[13] = 0.000011677472938534505691
 r[14] = 0.000005213671069071438316
 r[15] = 0.000002327732557563864794
 r[16] = 0.000001039250316156727364
 r[17] = 0.000000463987405298168625
 r[18] = 0.000000207153247753330111
 r[19] = 0.000000092486232892226818
 r[20] = 0.000000041291659743219714
 r[21] = 0.000000018435187011819875
 r[22] = 0.000000008230623510745206
 r[23] = 0.000000003674666388018384
 r[24] = 0.000000001640601459821044
 r[25] = 0.000000000732467346803569
 r[26] = 0.000000000327019344034464
 r[27] = 0.000000000146001936820081
 r[28] = 0.000000000065184417792960
 r[29] = 0.000000000029102410659074
 r[30] = 0.000000000012993140612251
 r[31] = 0.000000000005800952537653
 r[32] = 0.000000000002589908887149
 r[33] = 0.000000000001156297694242
 r[34] = 0.000000000000516243781524
 r[35] = 0.000000000000230483588516
 r[36] = 0.000000000000102902323430
 r[37] = 0.000000000000045942048350
 r[38] = 0.000000000000020511410590
 r[39] = 0.000000000000009157579592
 r[40] = 0.000000000000004088517639
 r[41] = 0.000000000000001825370592
 r[42] = 0.000000000000000814959869
 r[43] = 0.000000000000000363849177
 r[44] = 0.000000000000000162445083
 r[45] = 0.000000000000000072525669
 r[46] = 0.000000000000000032380005
 r[47] = 0.000000000000000014456464
 r[48] = 0.000000000000000006454272
 r[49] = 0.000000000000000002881592
 r[50] = 0.000000000000000001286523
 ...

[quote=ukster post_id=1269968 time=1763119758 user_id=76233]
... Bepaal de straal van de rode cirkel ...
[/quote]
[itex]r_{rood} = 0.0637486875121543961164957699218777512303920702311138759936813114925...[/itex]
[attachment=0]sctfilling.png[/attachment]


En zo doorgaand met cirkels [itex]C_i[/itex] met straal [itex]r_i[/itex]:

[code]
r[1] = 0.107206358681917884414824
r[2] = 0.063748687512154396116496
r[3] = 0.032821511636756146746755
r[4] = 0.015670122563985885102203
r[5] = 0.007214141551854388718512
r[6] = 0.003265777217972012640372
r[7] = 0.001467139761780204657812
r[8] = 0.000656847751170706173659
r[9] = 0.000293622807019700485942
r[10] = 0.000131164440584095479382
r[11] = 0.000058574551480610704541
r[12] = 0.000026154247424269826999
r[13] = 0.000011677472938534505691
r[14] = 0.000005213671069071438316
r[15] = 0.000002327732557563864794
r[16] = 0.000001039250316156727364
r[17] = 0.000000463987405298168625
r[18] = 0.000000207153247753330111
r[19] = 0.000000092486232892226818
r[20] = 0.000000041291659743219714
r[21] = 0.000000018435187011819875
r[22] = 0.000000008230623510745206
r[23] = 0.000000003674666388018384
r[24] = 0.000000001640601459821044
r[25] = 0.000000000732467346803569
r[26] = 0.000000000327019344034464
r[27] = 0.000000000146001936820081
r[28] = 0.000000000065184417792960
r[29] = 0.000000000029102410659074
r[30] = 0.000000000012993140612251
r[31] = 0.000000000005800952537653
r[32] = 0.000000000002589908887149
r[33] = 0.000000000001156297694242
r[34] = 0.000000000000516243781524
r[35] = 0.000000000000230483588516
r[36] = 0.000000000000102902323430
r[37] = 0.000000000000045942048350
r[38] = 0.000000000000020511410590
r[39] = 0.000000000000009157579592
r[40] = 0.000000000000004088517639
r[41] = 0.000000000000001825370592
r[42] = 0.000000000000000814959869
r[43] = 0.000000000000000363849177
r[44] = 0.000000000000000162445083
r[45] = 0.000000000000000072525669
r[46] = 0.000000000000000032380005
r[47] = 0.000000000000000014456464
r[48] = 0.000000000000000006454272
r[49] = 0.000000000000000002881592
r[50] = 0.000000000000000001286523
...
[/code]

Deze is net iets simpeler!
Twee cirkels, een vierkant en een kwart cirkel, zijn ingeschreven in een kwart cirkel met straal 1.
Bepaal de straal van de rode cirkel.
[attachment=0]1.png[/attachment]

[attachment=0]tweecirkels2.png[/attachment]

Alternatief zonder de afstandsformule:

[itex]\angle FEO = \angle COM[/itex]

dus

[itex]\frac{FO}{EO} = \frac{CM}{MO}[/itex]

ofwel

[itex]\frac{4}{\sqrt{80}} = \frac{R}{Y}[/itex]

ofwel

[itex]R = \frac{Y}{\sqrt{5}}[/itex]

en verder als hierboven.

[attachment=0]tweecirkels.png[/attachment]

Leg het vierkant in een assenstelsel met het punt rechtsonder = B = (4, 0)
Definieer:
lijn [itex]l: 2x + y = 0[/itex]
middelpunt blauwe cirkel = [itex]M = (0, Y)[/itex]
straal blauwe cirkel = [itex]R[/itex]
middelpunt rode cirkel = [itex]N[/itex]
straal rode cirkel = [itex]r[/itex]

Dan is:
[itex]d(M, l) = R = \frac{Y}{\sqrt{5}}[/itex]
(zie bv. [url]https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line[/url])
en
[itex]AB = 8 = R + BM = R + \sqrt{4^2 + Y^2}[/itex]
dus
[itex]\frac{Y}{\sqrt{5}} + \sqrt{4^2 + Y^2} = 8[/itex]
waardoor
[itex]Y = 2\sqrt{5}[/itex]
en
[itex]R = 2[/itex]

Gebruik dan de gelijkvormigheid van de (rechthoekige) driehoeken MCO en NDO:
[itex]\frac{OM}{MC}=\frac{ON}{ND}[/itex]
ofwel:
[itex]\frac{Y}{R} = \frac{Y-R-r}{r}[/itex]

[itex]Yr = YR-R^2-rR[/itex]

[itex](Y+R)r = YR-R^2[/itex]

[itex]r = \frac{YR-R^2}{Y+R}[/itex]

invullen: [itex]r = 3-\sqrt{5}[/itex]

ziet er iemand een oplossing om op basis van de gegevens in bijlage de straal van de kleine cirkel te berekenen