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Question-186616

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Question Number 186616 by ajfour last updated on 07/Feb/23
Answered by ajfour last updated on 07/Feb/23
say m=1 & forget the tangent. C(h,k)≡(1+p, r) let circle touches cubic curve at Q(q, q^3 −q−c) ⇒ (1+p−q)^2 +(q^3 −q−c−r)^2 =r^2 further, 3q^2 −1=((1+p−q)/(q^3 −q−c−r)) ⇒ p=q−1+(3q^2 −1)(q^3 −q−c−r) ⇒ {1+(3q^2 −1)^2 }(q^3 −q−c−r)^2 =r^2 lrt 1+(3q^2 −1)^2 =t^2 t(q^3 −q−c−r)=r p=q−1+r q=p+1−r 1+{3(p+1−r)^2 −1}^2 =t^2 ⇒(p+1−r)^2 =(1/3)±(1/3)(√(t^2 −1)) p^2 +2(1−r)p+(1−r)^2 =T p+c+2(1−r)p^2 +p(1−r)^2 =pT & 4(1−r)^2 p+2(1−r)p^2 +2(1−r)^3 =2(1−r)T subtracting {3(1−r)^2 −1+T}p =2(1−r)T+c−2(1−r)^3 ⇒ {2(1−r)T+c−2(1−r)^3 }^2 +2(1−r){2(1−r)T+c−2(1−r)^3 } ×{3(1−r)^2 −1+T} ={T−(1−r)^2 }{3(1−r)^2 −1+T}^2 say 1−r=R {2RT+c−2R^3 }^2 +2R{2RT+c−2R^3 }{T+3R^2 −1} =(T−R^2 )(T+3R^2 −1)^2 ⇒ T^( 3) −R^2 T^( 2) +2(3R^2 −1)T^( 2) −2R^2 (3R^2 −1)T+(3R^2 −1)^2 T −R^2 (3R^2 −1)^2 =4R^2 T^( 2) −4R(c−2R^3 )T+(c−2R^3 )^2 +4R^2 T^( 2) +4R^2 (3R^2 −1)T +2R(c−2R^3 )T+2R(c−2R^3 )(3R^2 −1) ⇒ if T=z, R=s z^3 −(3s^2 +2)z^2 +(1+2cs−13s^4 )z −{c−2s^3 +s(3s−1)}^2 =0 but 2s^3 −3s^2 +s−c=0 gives, for c=(1/3) s≈1.1989 now z^2 −(3s^2 +2)z+(1+2cs−13s^4 )=0 z=1+((3s^2 )/2)±(√((1+((3s^2 )/2))^2 +13s^4 −2cs−1)) p=r−1±(√T)=−s±(√z) .....
$${say}\:{m}=\mathrm{1}\:\&\:{forget}\:{the}\:{tangent}. \\ $$$${C}\left({h},{k}\right)\equiv\left(\mathrm{1}+{p},\:{r}\right) \\ $$$${let}\:{circle}\:{touches}\:{cubic}\:{curve}\:{at} \\ $$$${Q}\left({q},\:{q}^{\mathrm{3}} −{q}−{c}\right) \\ $$$$\Rightarrow\:\:\left(\mathrm{1}+{p}−{q}\right)^{\mathrm{2}} +\left({q}^{\mathrm{3}} −{q}−{c}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${further}, \\ $$$$\mathrm{3}{q}^{\mathrm{2}} −\mathrm{1}=\frac{\mathrm{1}+{p}−{q}}{{q}^{\mathrm{3}} −{q}−{c}−{r}} \\ $$$$\Rightarrow\:\:{p}={q}−\mathrm{1}+\left(\mathrm{3}{q}^{\mathrm{2}} −\mathrm{1}\right)\left({q}^{\mathrm{3}} −{q}−{c}−{r}\right) \\ $$$$\Rightarrow\:\:\left\{\mathrm{1}+\left(\mathrm{3}{q}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \right\}\left({q}^{\mathrm{3}} −{q}−{c}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${lrt}\:\:\:\:\mathrm{1}+\left(\mathrm{3}{q}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} ={t}^{\mathrm{2}} \\ $$$${t}\left({q}^{\mathrm{3}} −{q}−{c}−{r}\right)={r} \\ $$$${p}={q}−\mathrm{1}+{r} \\ $$$${q}={p}+\mathrm{1}−{r} \\ $$$$\mathrm{1}+\left\{\mathrm{3}\left({p}+\mathrm{1}−{r}\right)^{\mathrm{2}} −\mathrm{1}\right\}^{\mathrm{2}} ={t}^{\mathrm{2}} \\ $$$$\Rightarrow\left({p}+\mathrm{1}−{r}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}\pm\frac{\mathrm{1}}{\mathrm{3}}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}} \\ $$$${p}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{1}−{r}\right){p}+\left(\mathrm{1}−{r}\right)^{\mathrm{2}} ={T} \\ $$$${p}+{c}+\mathrm{2}\left(\mathrm{1}−{r}\right){p}^{\mathrm{2}} +{p}\left(\mathrm{1}−{r}\right)^{\mathrm{2}} ={pT} \\ $$$$\& \\ $$$$\mathrm{4}\left(\mathrm{1}−{r}\right)^{\mathrm{2}} {p}+\mathrm{2}\left(\mathrm{1}−{r}\right){p}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{1}−{r}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{1}−{r}\right){T} \\ $$$${subtracting} \\ $$$$\left\{\mathrm{3}\left(\mathrm{1}−{r}\right)^{\mathrm{2}} −\mathrm{1}+{T}\right\}{p} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{1}−{r}\right){T}+{c}−\mathrm{2}\left(\mathrm{1}−{r}\right)^{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$$\left\{\mathrm{2}\left(\mathrm{1}−{r}\right){T}+{c}−\mathrm{2}\left(\mathrm{1}−{r}\right)^{\mathrm{3}} \right\}^{\mathrm{2}} \\ $$$$\:\:+\mathrm{2}\left(\mathrm{1}−{r}\right)\left\{\mathrm{2}\left(\mathrm{1}−{r}\right){T}+{c}−\mathrm{2}\left(\mathrm{1}−{r}\right)^{\mathrm{3}} \right\} \\ $$$$\:\:\:\:\:\:\:×\left\{\mathrm{3}\left(\mathrm{1}−{r}\right)^{\mathrm{2}} −\mathrm{1}+{T}\right\} \\ $$$$\:\:\:\:\:=\left\{{T}−\left(\mathrm{1}−{r}\right)^{\mathrm{2}} \right\}\left\{\mathrm{3}\left(\mathrm{1}−{r}\right)^{\mathrm{2}} −\mathrm{1}+{T}\right\}^{\mathrm{2}} \\ $$$${say}\:\:\mathrm{1}−{r}={R} \\ $$$$\left\{\mathrm{2}{RT}+{c}−\mathrm{2}{R}^{\mathrm{3}} \right\}^{\mathrm{2}} \\ $$$$+\mathrm{2}{R}\left\{\mathrm{2}{RT}+{c}−\mathrm{2}{R}^{\mathrm{3}} \right\}\left\{{T}+\mathrm{3}{R}^{\mathrm{2}} −\mathrm{1}\right\} \\ $$$$\:\:\:\:=\left({T}−{R}^{\mathrm{2}} \right)\left({T}+\mathrm{3}{R}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${T}^{\:\mathrm{3}} −{R}^{\mathrm{2}} {T}^{\:\mathrm{2}} +\mathrm{2}\left(\mathrm{3}{R}^{\mathrm{2}} −\mathrm{1}\right){T}^{\:\mathrm{2}} \\ $$$$−\mathrm{2}{R}^{\mathrm{2}} \left(\mathrm{3}{R}^{\mathrm{2}} −\mathrm{1}\right){T}+\left(\mathrm{3}{R}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} {T} \\ $$$$\:\:\:−{R}^{\mathrm{2}} \left(\mathrm{3}{R}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\mathrm{4}{R}^{\mathrm{2}} {T}^{\:\mathrm{2}} −\mathrm{4}{R}\left({c}−\mathrm{2}{R}^{\mathrm{3}} \right){T}+\left({c}−\mathrm{2}{R}^{\mathrm{3}} \right)^{\mathrm{2}} \\ $$$$\:\:+\mathrm{4}{R}^{\mathrm{2}} {T}^{\:\mathrm{2}} +\mathrm{4}{R}^{\mathrm{2}} \left(\mathrm{3}{R}^{\mathrm{2}} −\mathrm{1}\right){T} \\ $$$$+\mathrm{2}{R}\left({c}−\mathrm{2}{R}^{\mathrm{3}} \right){T}+\mathrm{2}{R}\left({c}−\mathrm{2}{R}^{\mathrm{3}} \right)\left(\mathrm{3}{R}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\Rightarrow\:\:{if}\:\:\:{T}={z},\:\:{R}={s} \\ $$$${z}^{\mathrm{3}} −\left(\mathrm{3}{s}^{\mathrm{2}} +\mathrm{2}\right){z}^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{2}{cs}−\mathrm{13}{s}^{\mathrm{4}} \right){z} \\ $$$$−\left\{{c}−\mathrm{2}{s}^{\mathrm{3}} +{s}\left(\mathrm{3}{s}−\mathrm{1}\right)\right\}^{\mathrm{2}} =\mathrm{0} \\ $$$${but}\:\:\mathrm{2}{s}^{\mathrm{3}} −\mathrm{3}{s}^{\mathrm{2}} +{s}−{c}=\mathrm{0} \\ $$$${gives},\:\:{for}\:{c}=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\:\:\:{s}\approx\mathrm{1}.\mathrm{1989} \\ $$$${now}\:\: \\ $$$${z}^{\mathrm{2}} −\left(\mathrm{3}{s}^{\mathrm{2}} +\mathrm{2}\right){z}+\left(\mathrm{1}+\mathrm{2}{cs}−\mathrm{13}{s}^{\mathrm{4}} \right)=\mathrm{0} \\ $$$${z}=\mathrm{1}+\frac{\mathrm{3}{s}^{\mathrm{2}} }{\mathrm{2}}\pm\sqrt{\left(\mathrm{1}+\frac{\mathrm{3}{s}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{13}{s}^{\mathrm{4}} −\mathrm{2}{cs}−\mathrm{1}} \\ $$$${p}={r}−\mathrm{1}\pm\sqrt{{T}}=−{s}\pm\sqrt{{z}} \\ $$$$….. \\ $$

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