GSPkp Fp capm"dt} T <\Read the associated Word File to see how Inversion Theory can be used to determine the radiit 8\/t\PA)GPKFind the radii of the internal circles to the given square of side length atupzu AACD t  Z`CD tK!DC t).J!DC tB__l_4____T_ ___d__, _ _ _T 1DC t`A`s` ```x`0` ` ` ` `|`D`_]CC Bt Dragging red points shows how the Original Shape is transformed into the Inverted Image. Dragging the black points shows where the inverted points originate. Notice that this is only one positioning of the Circle of Inversion and is not necessarily the most appropriate for ease of construction. Try Dragging either the Circumference or the Centre of the Circle of Inversion to find a better transformation for the inversion (Click on the Hint for help).tG  1An Example to illustrate the method of Inversionst:mN Move K->AA t%d9 Move J->Z Xt(?Drag'?'?o8jp Rq vRt$ !DC#C3?+7?t Pj! vaCCr5H>:4CCwA7]CC1DC?t^r Reset Circle of Inversionp Rq vRt,\  tWl m23 Radius Drag = Radius(Circle Drag) = tWl m20 Radius Drag = Radius(Circle Drag) = t m17 Radius Drag = Radius(Circle Drag) = ti~ m14 Radius Drag = Radius(Circle Drag) = t* m11 Radius Drag = Radius(Circle Drag) = t m8 Radius Drag = Radius(Circle Drag) = tm m5 Radius Drag = Radius(Circle Drag) = t\? DAS? c10!DC t4? CBC%ى>BC)C,>CC6}˷>BmCuCC tg$ Move K->D  t aoZ'?'?o8jOq _O^tlA!DCCC8t? tRAl!DYC!D0D?tRAk2CYCC0D?t+(rI@c2h~ CC D2 wh!DC#CF%5?F%5?t>+rc1(|ACC#CF%5?F%5?tqG? P9NDC t). F!DC t). E HCC tc Move J->F  tg m10JP = Distance(J to P) =  tC(Yx!DC9NDC?t+(r˖@c3l 5!DC#CF%5?F%5? t(u,  CC!DC?! t(t0.74 cmCCCC?! t(rRadius(Circle EF) = !DC!DC`w?  t(.qs  =?Radius {!:CCC!DC? !tG[ Circle of Inversion (Hint))d d 3)/O "t-_? Roooo____￿￿ 00!D!C (t+RT m12 "{D:{(:Radius {!:C}JK}{u:2}}{JP} = 'Radius(Circle JK)^2/Distance(J to P) = #tx}2H? NdUC4C 't).> LDC )tY^ GLhv/D^C &%ta' a= a g m {S:FD} = a =  (tSgh m13````0000oooJR = Distance(J to R) = + t(Y`yoooo000?ߐ?oooo!DC!D!C?+t).+J'*ADC ,tg m7JN = Distance(J to N) = - t (Jw(S{Q{aP{{O{AN{CM{L{ NLpK{XK{J{mK{ L{JL{NM{O{S{!DCC4C?-tWel m4JL = Distance(J to L) = . t (.v!DCDC?.t@Wp m3  {(:3 - 2{!:*}{V:2}}{!:*}a = (3 - 2*sqrt[2] )*a =  50t+ a1!DC#C55!DCCC/ t(?X m2  {(:{V:2} - 1}{!:*}a = (sqrt[2] - 1)*a =  50t?!T]CMC 9t m15 "{D:{(:Radius {!:C}JK}{u:2}}{JR} = 'Radius(Circle JK)^2/Distance(J to R) = 1tc11!DCCF%5?F%5?3t m9 "{D:{(:Radius {!:C}JK}{u:2}}{JN} = 'Radius(Circle JK)^2/Distance(J to N) = 4t8 m6 "{D:{(:Radius {!:C}JK}{u:2}}{JL} = 'Radius(Circle JK)^2/Distance(J to L) = 6t+Y0^+G'LhO- D^C /8tRYW^+G'LhD^C /: tf(Yz!DC]CMC?;te m16JT = Distance(J to T) = ;t=)B.+J'ODC <t38 Q~@ DGC $=t).+J'CEDC >t~).+J'j`DC ?t?.xc61"sGP2xXv/D^C AF%5?F%5?/@tUc41"sGP2v/D^CBF%5?F%5?/Ato m18 "{D:{(:Radius {!:C}JK}{u:2}}{JT} = 'Radius(Circle JK)^2/Distance(J to T) = Ctp@c12!DCx:CF%5?F%5?D't+u( ?QE($#,3=Etc10!DC CF%5?F%5?Ft/(c9!DC|CF%5?F%5?GtEJS I [NZTsxX{C{C &Ht>C HD=C &It).+J'\$EDC Jt S!DFMS't$3( ?M`emTrc15!DCH>mCF%5?F%5?gtc14!DCCF%5?F%5?htEJ YQ9CQC `it#( WCC bj't$( ?Yk(UY`_dgik'tu( ?Wl(VZbaehjlt+? Hide Inverted Image>L:EN@$܈mg((g{ l fm^]Ln\t* Show Inverted Image>L:EN@$܈mg((g{ l  fm^]Ln\"ArialGO4X