Question Number 101375 by ajfour last updated on 02/Jul/20
Commented by ajfour last updated on 02/Jul/20
$${In}\:{terms}\:{of}\:{R},\:{find}\:{radii}\:{a},{b},{c}. \\ $$
Commented by ajfour last updated on 05/Jul/20
Commented by ajfour last updated on 06/Jul/20
$${lets}\:{consider}\:{both}\:{circles}\:{of}\:{radii} \\ $$$${a}\:{and}\:{c}\:{generally}. \\ $$$${Both}\:{are}\:{at}\:{a}\:{distance}\:{of}\:\left({R}+{r}\right)\:{from} \\ $$$${center}\:{M}\:{of}\:{circle}\:{of}\:{radius}\:{R}. \\ $$$${Both}\:{touch}\:{the}\:{line}\:{y}={R}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}+{R} \\ $$$${both}\:{touch}\:{the}\:{parabola}\:{at}\:\left({x},{x}^{\mathrm{2}} \right) \\ $$$${x}_{\mathrm{1}} =\sqrt{{R}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${center}\:{of}\:{our}\:{general}\:{circle}\:{is} \\ $$$$\:\:\:\:\:\left(\mathrm{2}\sqrt{{Rr}}\:,\:{R}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}+{R}−{r}\right) \\ $$$${eq}.\:{of}\:{the}\:{circle}: \\ $$$$\begin{cases}{\left({x}−\mathrm{2}\sqrt{{Rr}}\right)^{\mathrm{2}} +\left[{x}^{\mathrm{2}} −\left({R}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}+{R}−{r}\right)\right]^{\mathrm{2}} ={r}^{\mathrm{2}} }\\{\left({R}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −{r}−\frac{{r}}{\:\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }}\:=\:{x}^{\mathrm{2}} }\end{cases} \\ $$$$\:\:\:\:{and}\:\:\:{x}−\mathrm{2}\sqrt{{Rr}}\:=\:\frac{{r}\left(\mathrm{2}{x}\right)}{\:\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }}\:\:\:{for}\:{r}={c} \\ $$$${but}\:\:\:\:\:\mathrm{2}\sqrt{{Rr}}−{x}=\frac{{r}\left(\mathrm{2}{x}\right)}{\:\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }}\:\:\:\:\:\:\:{for}\:{r}={a} \\ $$$${And}\:{since}\:{we}\:{know}\:{x}_{\mathrm{1}} =\sqrt{{R}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\:\:\:\:\:\mathrm{2}\sqrt{{Rr}_{\mathrm{1}} }={x}+\frac{\mathrm{2}{x}_{\mathrm{1}} {r}}{\:\sqrt{\mathrm{1}+\mathrm{4}{x}_{\mathrm{1}} ^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\mathrm{2}\sqrt{{Ra}}=\sqrt{{R}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}\:+\:\frac{\mathrm{2}{a}\sqrt{{R}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}}{\:\sqrt{\mathrm{1}+\mathrm{4}{R}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\mathrm{2}\sqrt{{Ra}}=\sqrt{{R}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}\:+\:\frac{{a}}{{R}}\sqrt{{R}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\frac{{a}}{{R}}\sqrt{{R}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}\:−\mathrm{2}{R}\sqrt{\frac{{a}}{{R}}}\:+\sqrt{{R}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}=\mathrm{0} \\ $$$$\sqrt{\frac{{a}}{{R}}}=\frac{{R}}{\:\sqrt{{R}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}}\pm\sqrt{\frac{{R}^{\mathrm{2}} }{{R}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\sqrt{\frac{{a}}{{R}}}\:=\frac{{R}\pm\frac{\mathrm{1}}{\mathrm{2}}}{\:\sqrt{{R}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}} \\ $$$$\Rightarrow\:\:\frac{{a}}{{R}}=\frac{\left({R}\pm\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{\left({R}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({R}−\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$${It}\:{seems}\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\frac{\boldsymbol{{a}}}{\boldsymbol{{R}}}=\frac{\left(\boldsymbol{{R}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left(\boldsymbol{{R}}−\frac{\mathrm{1}}{\mathrm{2}}\right)}\:\:\:\:\: \\ $$$${example}\:\:\:{R}=\mathrm{2}\:\:\Rightarrow\:\:\:{a}=\frac{\mathrm{10}}{\mathrm{3}} \\ $$$$\:……….\:\:\:\:\:\: \\ $$
Commented by ajfour last updated on 05/Jul/20
Answered by mr W last updated on 05/Jul/20
$${Center}\:{of}\:{circle}\:{with}\:{radius}\:{R}\:{is} \\ $$$${M}\left(\mathrm{0},{m}\right) \\ $$$${y}={x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\left({y}−{m}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow{y}+\left({y}−{m}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow{y}^{\mathrm{2}} −\left(\mathrm{2}{m}−\mathrm{1}\right){y}+{m}^{\mathrm{2}} −{R}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta=\left(\mathrm{2}{m}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left({m}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{1}+\mathrm{4}{R}^{\mathrm{2}} −\mathrm{4}{m}=\mathrm{0} \\ $$$$\Rightarrow{m}=\frac{\mathrm{1}}{\mathrm{4}}+{R}^{\mathrm{2}} \\ $$$${Center}\:{of}\:{circle}\:{with}\:{radius}\:{b}\:{is} \\ $$$${B}\left(\mathrm{0},{m}+{R}+{b}\right) \\ $$$${x}^{\mathrm{2}} +\left({y}−{m}−{R}−{b}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$${y}+\left[{y}−\left(\frac{\mathrm{1}}{\mathrm{4}}+{R}^{\mathrm{2}} +{R}+{b}\right)\right]^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} −\mathrm{2}\left({R}^{\mathrm{2}} +{R}+{b}−\frac{\mathrm{1}}{\mathrm{4}}\right){y}+\left({R}^{\mathrm{2}} +{R}+{b}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta=\left({R}^{\mathrm{2}} +{R}+{b}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} −\left({R}^{\mathrm{2}} +{R}+{b}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${b}^{\mathrm{2}} −{b}−\left({R}^{\mathrm{2}} +{R}\right)=\mathrm{0} \\ $$$$\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{2}{R}+\mathrm{1}\right)={R}+\mathrm{1} \\ $$$$ \\ $$$${Center}\:{of}\:{circle}\:{with}\:{radius}\:{a}\:{is} \\ $$$${A}\left(\mathrm{2}\sqrt{{Ra}},{m}+{R}−{a}\right) \\ $$$${it}\:{touches}\:{the}\:{parabola}\:{at}\:{P}\left({p},\:{p}^{\mathrm{2}} \right). \\ $$$$\mathrm{tan}\:\theta=\mathrm{2}{p}=\frac{\mathrm{2}\sqrt{{Ra}}}{{a}−{R}} \\ $$$$\Rightarrow{p}=\frac{\sqrt{{Ra}}}{{a}−{R}} \\ $$$$\left(\mathrm{2}\sqrt{{Ra}}−{p}\right)^{\mathrm{2}} +\left({m}+{R}−{a}−{p}^{\mathrm{2}} \right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\sqrt{{Ra}}−\frac{\sqrt{{Ra}}}{{a}−{R}}\right)^{\mathrm{2}} +\left[{m}+{R}−{a}−\frac{{Ra}}{\left({a}−{R}\right)^{\mathrm{2}} }\right]^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\left[\mathrm{2}−\frac{\mathrm{1}}{{R}\left(\frac{{a}}{{R}}−\mathrm{1}\right)}\right]^{\mathrm{2}} \left(\frac{{a}}{{R}}\right)+\frac{\mathrm{1}}{{R}^{\mathrm{2}} }\left[{m}−{R}\left(\frac{{a}}{{R}}−\mathrm{1}\right)−\frac{\frac{{a}}{{R}}}{\left(\frac{{a}}{{R}}−\mathrm{1}\right)^{\mathrm{2}} }\right]^{\mathrm{2}} =\left(\frac{{a}}{{R}}\right)^{\mathrm{2}} \\ $$$$\left[\mathrm{2}−\frac{\mathrm{1}}{{R}\left(\xi−\mathrm{1}\right)}\right]^{\mathrm{2}} \xi+\frac{\mathrm{1}}{{R}^{\mathrm{2}} }\left[\frac{\mathrm{1}}{\mathrm{4}}+{R}^{\mathrm{2}} −{R}\left(\xi−\mathrm{1}\right)−\frac{\xi}{\left(\xi−\mathrm{1}\right)^{\mathrm{2}} }\right]^{\mathrm{2}} =\xi^{\mathrm{2}} \\ $$$$\Rightarrow\xi=… \\ $$$$ \\ $$$${Center}\:{of}\:{circle}\:{with}\:{radius}\:{c}\:{is} \\ $$$${C}\left(\mathrm{2}\sqrt{{Rc}},{m}+{R}−{c}\right) \\ $$$${it}\:{touches}\:{the}\:{parabola}\:{at}\:{Q}\left({q},\:{q}^{\mathrm{2}} \right). \\ $$$$\mathrm{tan}\:\varphi=\mathrm{2}{q} \\ $$$${q}=\mathrm{2}\sqrt{{Rc}}+{c}\:\mathrm{sin}\:\varphi=\mathrm{2}\sqrt{{Rc}}+\frac{\mathrm{2}{cq}}{\:\sqrt{\mathrm{1}+\mathrm{4}{q}^{\mathrm{2}} }} \\ $$$$\Rightarrow{q}−\mathrm{2}\sqrt{{Rc}}=\frac{\mathrm{2}{cq}}{\:\sqrt{\mathrm{1}+\mathrm{4}{q}^{\mathrm{2}} }}\:\:\:…\left({i}\right) \\ $$$${q}^{\mathrm{2}} ={m}+{R}−{c}−{c}\:\mathrm{cos}\:\varphi={m}+{R}−{c}−\frac{{c}}{\:\sqrt{\mathrm{1}+\mathrm{4}{q}^{\mathrm{2}} }} \\ $$$$\Rightarrow{m}+{R}−{c}−{q}^{\mathrm{2}} =\frac{{c}}{\:\sqrt{\mathrm{1}+\mathrm{4}{q}^{\mathrm{2}} }}\:\:\:\:…\left({ii}\right) \\ $$$${q}−\mathrm{2}\sqrt{{Rc}}=\mathrm{2}{q}\left({m}+{R}−{c}−{q}^{\mathrm{2}} \right) \\ $$$${q}^{\mathrm{3}} −\left({R}^{\mathrm{2}} +{R}−{c}−\frac{\mathrm{1}}{\mathrm{4}}\right){q}−\sqrt{{Rc}}=\mathrm{0} \\ $$$$\Rightarrow{q}=\sqrt[{\mathrm{3}}]{\frac{\sqrt{{Rc}}}{\mathrm{2}}+\sqrt{\frac{{Rc}}{\mathrm{4}}−\frac{\left({R}^{\mathrm{2}} +{R}−{c}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{3}} }{\mathrm{27}}}}+\sqrt[{\mathrm{3}}]{\frac{\sqrt{{Rc}}}{\mathrm{2}}−\sqrt{\frac{{Rc}}{\mathrm{4}}−\frac{\left({R}^{\mathrm{2}} +{R}−{c}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{3}} }{\mathrm{27}}}} \\ $$$${put}\:{this}\:{into}\:\left({i}\right)\:{or}\:\left({ii}\right)\:{to}\:{get}\:{c} \\ $$
Commented by mr W last updated on 05/Jul/20
Commented by ajfour last updated on 05/Jul/20
$${but}\:{c}\:{is}\:{to}\:{be}\:{determined},\:{Sir}. \\ $$
Commented by mr W last updated on 05/Jul/20
$${a}\:{and}\:{c}\:{can}\:{only}\:{be}\:{determined}\: \\ $$$${numerically}. \\ $$
Commented by ajfour last updated on 05/Jul/20
$${if}\:{we}\:{discuss},\:{i}\:{think}\:{we}\:{can}\:{get}\:{the} \\ $$$${expression}\:{for}\:{radii}\:{a}\:{and}\:{c}\:,\:{Sir}.. \\ $$
Commented by mr W last updated on 05/Jul/20
$${i}\:{found}\:{no}\:{way}\:{to}\:{get}\:{them}. \\ $$
Commented by ajfour last updated on 06/Jul/20
$$\:\:\:\:{Sir}\:{i}\:{got}\:\:{a}={R}\left({R}+\mathrm{1}/\mathrm{2}\right)/\left({R}−\mathrm{1}/\mathrm{2}\right) \\ $$$${Expression}\:{for}\:{c}\:{indeed}\:{seems} \\ $$$${difficult}. \\ $$
Commented by mr W last updated on 06/Jul/20
$${you}\:{got}\:{it}\:{at}\:{least}\:{for}\:{a}!\:{it}'{s}\:{correct}. \\ $$