Question-225856

Question Number 225856 by ajfour last updated on 15/Nov/25

Commented by ajfour last updated on 15/Nov/25

$${Parabola}\:{shown}\:{is}\:\:{y}={x}^{\mathrm{2}} . \\ $$$${Find}\:{equations}\:{of}\:{maximum} \\ $$$${radius}\:{yellow}\:{circles},\:{or}\:{find} \\ $$$$\:\:\:\:\:\:\:{r},\:{C}\left({h},{k}\right)\:,\:{C}\:'\left(−{h},{k}\right) \\ $$

Commented by mr W last updated on 15/Nov/25

$${radius}\:{of}\:{blue}\:{circle}\:{must}\:{be}\:{given}. \\ $$

Commented by ajfour last updated on 15/Nov/25

$${Take}\:{it}\:{just}\:{R}.\:{as}\:{an}\:{example} \\ $$$${take}\:{R}=\mathrm{2}. \\ $$

Commented by mr W last updated on 15/Nov/25

Commented by mr W last updated on 15/Nov/25

Commented by mr W last updated on 15/Nov/25

Commented by mr W last updated on 15/Nov/25

$${we}\:{can}\:{only}\:{solve}\:{for}\:{r}\:{nummerically}. \\ $$

Commented by ajfour last updated on 15/Nov/25

$${I}\:{got}\:{r}=\frac{\mathrm{1}}{\mathrm{2}}\left({R}−\sqrt{{R}−\frac{\mathrm{1}}{\mathrm{4}}}\right) \\ $$$${for}\:{R}=\mathrm{2}\:{eq}\:{of}\:{right}\:{yellow}\:{circle}\:{is} \\ $$$$\left({x}−\frac{\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{7}}}−\frac{\sqrt{\mathrm{6}}}{\mathrm{4}}\right)^{\mathrm{2}} +\left({y}−\frac{\mathrm{7}}{\mathrm{4}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\right)^{\mathrm{2}} =\left(\mathrm{1}−\frac{\sqrt{\mathrm{7}}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$

Commented by ajfour last updated on 15/Nov/25

Commented by fantastic2 last updated on 15/Nov/25

$${i}\:{think}\: \\ $$$$\begin{cases}{\left({x}−{h}\right)^{\mathrm{2}} +\left({y}−{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} }\\{{y}={x}^{\mathrm{2}} }\end{cases}\Rightarrow{x}_{\mathrm{1}} =\alpha\:{y}_{\mathrm{1}} =\beta \\ $$$$\alpha\:\&\beta\:{will}\:{be}\:{in}\:{terms}\:{of}\:{k},{h\&r} \\ $$$$\begin{cases}{\left({x}−{h}\right)^{\mathrm{2}} +\left({y}−{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} }\\{{x}^{\mathrm{2}} +\left({y}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} }\end{cases}\Rightarrow{x}_{\mathrm{2}} =\gamma\:{y}_{\mathrm{2}} =\delta \\ $$$$\gamma\&\delta\:{will}\:{be}\:{in}\:{terms}\:{of}\:{R},{h\&k} \\ $$$${at}\:{r}_{{max}} \\ $$$$\sqrt{\left({x}_{\mathrm{1}} −{x}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({y}_{\mathrm{1}} −{y}_{\mathrm{2}} \right)^{\mathrm{2}} }\:{will}\:{be}\:{maximum} \\ $$

Commented by mr W last updated on 15/Nov/25

$${ajfour}\:{sir}: \\ $$$${you}\:{are}\:{right}\:{sir}! \\ $$$${great}\:{job}! \\ $$

Commented by ajfour last updated on 15/Nov/25

I agree tobyour reasoning. see my comment solution.

Answered by mr W last updated on 15/Nov/25

$${this}\:{is}\:{my}\:{way}: \\ $$$${y}=\frac{{x}^{\mathrm{2}} }{{c}}\:{with}\:{c}=\mathrm{1} \\ $$$${say}\:{small}\:{yellow}\:{circle}\:{touches}\:{the} \\ $$$${parabola}\:{at}\:\left({p},\:\frac{{p}^{\mathrm{2}} }{{c}}\right). \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2}{p}}{{c}} \\ $$$${h}={p}+{r}\:\mathrm{sin}\:\theta={p}+\frac{\mathrm{2}{pr}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{c}}} \\ $$$${k}=\frac{{p}^{\mathrm{2}} }{{c}}−{r}\:\mathrm{cos}\:\theta=\frac{{p}^{\mathrm{2}} }{{c}}−\frac{{cr}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{c}^{\mathrm{2}} }} \\ $$$${since}\:{small}\:{yellow}\:{circle}\:{tangents} \\ $$$${the}\:{big}\:{blue}\:{circle}, \\ $$$$\left({p}+\frac{\mathrm{2}{pr}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{c}}}\right)^{\mathrm{2}} +\left({R}−\frac{{p}^{\mathrm{2}} }{{c}}+\frac{{cr}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{c}^{\mathrm{2}} }}\right)^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} \\ $$$${p}^{\mathrm{2}} +\frac{\mathrm{4}{p}^{\mathrm{2}} {r}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{c}}}+\frac{\mathrm{4}{p}^{\mathrm{2}} {r}^{\mathrm{2}} }{\mathrm{4}{p}^{\mathrm{2}} +{c}^{\mathrm{2}} }+\left({R}−\frac{{p}^{\mathrm{2}} }{{c}}\right)^{\mathrm{2}} +\mathrm{2}\left({R}−\frac{{p}^{\mathrm{2}} }{{c}}\right)\frac{{cr}}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{c}^{\mathrm{2}} }}+\frac{{c}^{\mathrm{2}} {r}^{\mathrm{2}} }{\mathrm{4}{p}^{\mathrm{2}} +{c}^{\mathrm{2}} }={R}^{\mathrm{2}} −\mathrm{2}{Rr}+{r}^{\mathrm{2}} \\ $$$$\mathrm{2}\left[\frac{\left({cR}+{p}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{c}^{\mathrm{2}} }}+{R}\right]{r}={p}^{\mathrm{2}} \left(\frac{\mathrm{2}{R}}{{c}}−\mathrm{1}−\frac{{p}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{r}=\frac{{p}^{\mathrm{2}} \left(\frac{\mathrm{2}{R}}{{c}}−\mathrm{1}−\frac{{p}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\right)}{\mathrm{2}\left({R}+\frac{{cR}+{p}^{\mathrm{2}} }{\:\sqrt{\mathrm{4}{p}^{\mathrm{2}} +{c}^{\mathrm{2}} }}\right)} \\ $$$$\frac{{r}}{{c}}=\frac{\frac{{p}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\left(\frac{\mathrm{2}{R}}{{c}}−\mathrm{1}−\frac{{p}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\right)}{\mathrm{2}\left(\frac{{R}}{{c}}+\frac{\frac{{R}}{{c}}+\frac{{p}^{\mathrm{2}} }{{c}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}+\frac{\mathrm{4}{p}^{\mathrm{2}} }{{c}^{\mathrm{2}} }}}\right)} \\ $$$${with}\:\xi=\frac{{p}^{\mathrm{2}} }{{c}^{\mathrm{2}} },\:\lambda=\frac{{R}}{{c}} \\ $$$$\frac{\mathrm{2}{r}}{{c}}=\:=\frac{\xi\left(\mathrm{2}\lambda−\mathrm{1}−\xi\right)}{\lambda+\frac{\lambda+\xi}{\:\sqrt{\mathrm{1}+\mathrm{4}\xi}}} \\ $$$${for}\:{maximum}\:{radius}\:{r}:\:\frac{{d}\:}{{d}\xi}=\mathrm{0}. \\ $$

Commented by mr W last updated on 15/Nov/25

$$\underline{{gravitation}\:{question}} \\ $$$${F}=\frac{{GMm}}{{x}^{\mathrm{2}} }={mA} \\ $$$${A}=\frac{{GM}}{{x}^{\mathrm{2}} }=\frac{{dv}}{{dt}}=−{v}\frac{{dv}}{{dx}} \\ $$$${vdv}=−\frac{{GMdx}}{{x}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{{v}} {vdv}=−\int_{{a}} ^{{x}} \frac{{GMdx}}{{x}^{\mathrm{2}} } \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}={GM}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{a}}\right) \\ $$$${v}=−\frac{{dx}}{{dt}}=\sqrt{\mathrm{2}{GM}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{a}}\right)} \\ $$$${dt}=−\frac{{dx}}{\:\sqrt{\mathrm{2}{GM}}\sqrt{\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{a}}}} \\ $$$$\int_{\mathrm{0}} ^{{T}} {dt}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{GM}}}\int_{{a}} ^{{R}} \frac{{dx}}{\:\sqrt{\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{a}}}} \\ $$$${T}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{GM}}}\int_{{a}} ^{{R}} \frac{\sqrt{{ax}}{dx}}{\:\sqrt{{a}−{x}}} \\ $$$${T}=\frac{\sqrt{{a}}}{\:\sqrt{\mathrm{2}{GM}}}\left[\sqrt{{x}\left({a}−{x}\right)}+{a}\:\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{{a}−{x}}{{x}}}\right]_{{a}} ^{{R}} \\ $$$${T}={a}\sqrt{\frac{{a}}{\mathrm{2}{GM}}}\left[\sqrt{\frac{{R}}{{a}}\left(\mathrm{1}−\frac{{R}}{{a}}\right)}+\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{{a}}{{R}}−\mathrm{1}}\right] \\ $$$$\Rightarrow{T}={a}\sqrt{\frac{{a}}{\mathrm{2}{GM}}}\left[\sqrt{\frac{{R}}{{a}}\left(\mathrm{1}−\frac{{R}}{{a}}\right)}+\mathrm{cos}^{−\mathrm{1}} \sqrt{\frac{{R}}{{a}}}\right] \\ $$

Commented by mr W last updated on 15/Nov/25

$${but}\:{we}\:{can}\:{short}\:{cut},\:{since}\:{for} \\ $$$${maximum}\:{yellow}\:{circle}\:{the}\:{tangent} \\ $$$${line}\:{on}\:{blue}\:{circle}\:{must}\:{parallel}\:{to} \\ $$$${the}\:{tangent}\:{line}\:{on}\:{the}\:{parabola}.\: \\ $$$${then}\:{we}\:{can}\:{solve}\:{directly}\:{for}\:{r}. \\ $$

Commented by ajfour last updated on 15/Nov/25

$${I}\:{got}\:{a}\:{simpler}\:{way} \\ $$$${let}\:{point}\:{of}\:{tangency}\:\left({p},\frac{{p}^{\mathrm{2}} }{{c}}\right) \\ $$$${eq}.\:{of}\:{tangent} \\ $$$${y}−\frac{{p}^{\mathrm{2}} }{{c}}=\mathrm{2}{p}\left({x}−{p}\right) \\ $$$${If}\:\boldsymbol{{s}}\:{shortest}\:{distance}\:{from}\:\left(\mathrm{0},{R}\right) \\ $$$${then} \\ $$$${s}^{\mathrm{2}} =\frac{\left({R}+\mathrm{2}{p}^{\mathrm{2}} −\frac{{p}^{\mathrm{2}} }{{c}}\right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} } \\ $$$$\frac{{d}\left({s}^{\mathrm{2}} \right)}{{dp}}=\mathrm{0}\:\:\Rightarrow\:\: \\ $$$$\mathrm{2}\left({R}+\mathrm{2}{p}^{\mathrm{2}} −\frac{{p}^{\mathrm{2}} }{{c}}\right)\left(\mathrm{4}{p}−\frac{\mathrm{2}{p}}{{c}}\right)\left(\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} \right) \\ $$$$\:\:=\mathrm{8}{p}\left({R}+\mathrm{2}{p}^{\mathrm{2}} −\frac{{p}^{\mathrm{2}} }{{c}}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{c}}\right)\left(\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} \right)={R}+\mathrm{2}{p}^{\mathrm{2}} −\frac{{p}^{\mathrm{2}} }{{c}} \\ $$$${p}^{\mathrm{2}} \left(\mathrm{4}−\frac{\mathrm{2}}{{c}}−\mathrm{2}+\frac{\mathrm{1}}{{c}}\right)={R}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{c}} \\ $$$${p}^{\mathrm{2}} =\frac{\left({R}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{c}}\right)}{\left(\mathrm{2}−\frac{\mathrm{1}}{{c}}\right)} \\ $$$${R}−{q}={R}−\frac{{p}^{\mathrm{2}} }{{c}}= \\ $$$${BC}^{\:\mathrm{2}} ={p}^{\mathrm{2}} +\left({R}−\frac{{p}^{\mathrm{2}} }{{c}}\right)^{\mathrm{2}} =\left({R}−\mathrm{2}{r}\right)^{\mathrm{2}} \\ $$$${For}\:\:{c}=\mathrm{1} \\ $$$${p}^{\mathrm{2}} ={R}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({R}−\mathrm{2}{r}\right)^{\mathrm{2}} =\left({R}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${r}=\frac{\mathrm{1}}{\mathrm{2}}\left({R}−\sqrt{{R}−\frac{\mathrm{1}}{\mathrm{4}}}\:\right) \\ $$

Commented by ajfour last updated on 15/Nov/25

$${I}\:{still}\:{have}\:{another}\:{way}. \\ $$$${but}\:{please}\:{watch}\:{my}\:{gravitation} \\ $$$${lecture}.\: \\ $$

Commented by ajfour last updated on 15/Nov/25

https://youtu.be/dH2KayLTr_M?si=S2RrzCitH7O6asVy

Commented by mr W last updated on 15/Nov/25

$${yes},\:{this}\:{is}\:{also}\:{what}\:{i}\:{have}\:{meant}. \\ $$

Commented by mr W last updated on 15/Nov/25

Commented by ajfour last updated on 15/Nov/25

$${yes}\:{sir}!\:{you}\:{got}\:{it}. \\ $$

Commented by fantastic2 last updated on 15/Nov/25

$${MrW}\:{sir}\::{reasonable} \\ $$

Answered by ajfour last updated on 15/Nov/25

$${To}\:{find}\:{maximum}\:{radius}\:{r}\:{of} \\ $$$${yellow}\:{circle},\:{we}\:{do}\:{this}. \\ $$$${We}\:{minimise}\:{distance}\:{from}\:{center} \\ $$$${of}\:{large}\:{circle}\:{of}\:{radiusR}. \\ $$$${say}\:{tangency}\:{point}\:{on}\:{parabola} \\ $$$${be}\:{P}\left({x},{x}^{\mathrm{2}} \right)\:{because}\:{our}\:{parabola} \\ $$$${equation}\:{is}\:\boldsymbol{{y}}=\boldsymbol{{x}}^{\mathrm{2}} . \\ $$$$\left({R}−\mathrm{2}{r}\right)^{\mathrm{2}} ={f}\left({x}\right)={x}^{\mathrm{2}} +\left({R}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${For}\:{maximum}\:{r},\:{we}\:{minimise} \\ $$$${f}\left({x}\right)\:{by}\:{setting}\:\frac{{df}\left({x}\right)}{{dx}}=\mathrm{0} \\ $$$${hence} \\ $$$$\mathrm{2}{x}+\mathrm{2}\left({R}−{x}^{\mathrm{2}} \right)\left(−\mathrm{2}{x}\right)=\mathrm{0} \\ $$$${as}\:{x}\neq\mathrm{0}\:{we}\:{obtain} \\ $$$${x}^{\mathrm{2}} ={R}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}=\sqrt{{R}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${now} \\ $$$$\left({R}−\mathrm{2}{r}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\left({R}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${we}\:{replace}\:{here}\:{found}\:{value}\:{of}\:{x}. \\ $$$$\left({R}−\mathrm{2}{r}\right)^{\mathrm{2}} ={R}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}={R}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${we}\:{clearly}\:{and}\:{easily}\:{now}\:{get} \\ $$$${r}=\frac{\mathrm{1}}{\mathrm{2}}\left({R}−\sqrt{{R}−\frac{\mathrm{1}}{\mathrm{4}}}\:\right) \\ $$$${And}\:{if}\:{our}\:{given}\:{R}=\mathrm{2}\:\:{then} \\ $$$${we}\:{plug}\:{in}\:{this}\:{value}\:{to}\:{get} \\ $$$${r}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}−\sqrt{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}}}\right) \\ $$$$\:\:\:\boldsymbol{{r}}=\mathrm{1}−\frac{\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$$$ \\ $$

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