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Question-219621

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Question Number 219621 by ajfour last updated on 29/Apr/25
Commented by ajfour last updated on 29/Apr/25
Correction: A(0,k^2 ) Find r in terms of k.
$${Correction}:\:{A}\left(\mathrm{0},{k}^{\mathrm{2}} \right) \\ $$$${Find}\:{r}\:{in}\:{terms}\:{of}\:{k}. \\ $$
Answered by mr W last updated on 29/Apr/25
touching point P(p,p^2 +k^2 ) tan θ=2p r=p+r sin θ=p+((2pr)/( (√(1+4p^2 )))) ⇒r=(p/(1−((2p)/( (√(1+4p^2 )))))) r=p^2 +k^2 −r cos θ=p^2 +k^2 −(r/( (√(1+4p^2 )))) r(1+(1/( (√(1+4p^2 )))))=p^2 +k^2 (p/( (√(1+4p^2 ))−2p))(1+(√(1+4p^2 )))=p^2 +k^2 ⇒p((√(1+4p^2 ))+2p)((√(1+4p^2 ))+1)=p^2 +k^2
$${touching}\:{point}\:{P}\left({p},{p}^{\mathrm{2}} +{k}^{\mathrm{2}} \right) \\ $$$$\mathrm{tan}\:\theta=\mathrm{2}{p} \\ $$$${r}={p}+{r}\:\mathrm{sin}\:\theta={p}+\frac{\mathrm{2}{pr}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$$\Rightarrow{r}=\frac{{p}}{\mathrm{1}−\frac{\mathrm{2}{p}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}} \\ $$$${r}={p}^{\mathrm{2}} +{k}^{\mathrm{2}} −{r}\:\mathrm{cos}\:\theta={p}^{\mathrm{2}} +{k}^{\mathrm{2}} −\frac{{r}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$${r}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}\right)={p}^{\mathrm{2}} +{k}^{\mathrm{2}} \\ $$$$\frac{{p}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }−\mathrm{2}{p}}\left(\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }\right)={p}^{\mathrm{2}} +{k}^{\mathrm{2}} \\ $$$$\Rightarrow{p}\left(\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }+\mathrm{2}{p}\right)\left(\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }+\mathrm{1}\right)={p}^{\mathrm{2}} +{k}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 29/Apr/25
Commented by mr W last updated on 29/Apr/25
i think r=f(k) as formula is not possible.
$${i}\:{think}\:{r}={f}\left({k}\right)\:{as}\:{formula}\:{is}\:{not} \\ $$$${possible}. \\ $$
Commented by mr W last updated on 30/Apr/25
r−p=r sin θ if center of circle is at (r,0), then p^2 +k^2 =r cos θ ((r−p)/(p^2 +k^2 ))=2p ⇒r=p(2p^2 +2k^2 +1) p^2 −2pr+p^4 +k^4 +2p^2 k^2 =0 3p^4 +(2k^2 +1)p^2 −k^4 =0 p^2 =(((√(16k^4 +4k^2 +1))−2k^2 −1)/6) ⇒p=(√(((√(16k^4 +4k^2 +1))−2k^2 −1)/6))
$${r}−{p}={r}\:\mathrm{sin}\:\theta \\ $$$${if}\:{center}\:{of}\:{circle}\:{is}\:{at}\:\left({r},\mathrm{0}\right),\:{then} \\ $$$${p}^{\mathrm{2}} +{k}^{\mathrm{2}} ={r}\:\mathrm{cos}\:\theta \\ $$$$\frac{{r}−{p}}{{p}^{\mathrm{2}} +{k}^{\mathrm{2}} }=\mathrm{2}{p} \\ $$$$\Rightarrow{r}={p}\left(\mathrm{2}{p}^{\mathrm{2}} +\mathrm{2}{k}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$ \\ $$$${p}^{\mathrm{2}} −\mathrm{2}{pr}+{p}^{\mathrm{4}} +{k}^{\mathrm{4}} +\mathrm{2}{p}^{\mathrm{2}} {k}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{3}{p}^{\mathrm{4}} +\left(\mathrm{2}{k}^{\mathrm{2}} +\mathrm{1}\right){p}^{\mathrm{2}} −{k}^{\mathrm{4}} =\mathrm{0} \\ $$$${p}^{\mathrm{2}} =\frac{\sqrt{\mathrm{16}{k}^{\mathrm{4}} +\mathrm{4}{k}^{\mathrm{2}} +\mathrm{1}}−\mathrm{2}{k}^{\mathrm{2}} −\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow{p}=\sqrt{\frac{\sqrt{\mathrm{16}{k}^{\mathrm{4}} +\mathrm{4}{k}^{\mathrm{2}} +\mathrm{1}}−\mathrm{2}{k}^{\mathrm{2}} −\mathrm{1}}{\mathrm{6}}} \\ $$
Answered by ajfour last updated on 30/Apr/25
Commented by ajfour last updated on 30/Apr/25
Commented by ajfour last updated on 30/Apr/25
Commented by ajfour last updated on 30/Apr/25
Commented by mr W last updated on 30/Apr/25
when the center of circle lies on x−axis, then it′s correct. but in the original question the circle should tangent both y−axis and x−axis.
$${when}\:{the}\:{center}\:{of}\:{circle}\:{lies}\:{on}\: \\ $$$${x}−{axis},\:{then}\:{it}'{s}\:{correct}.\:{but}\:{in}\:{the} \\ $$$${original}\:{question}\:{the}\:{circle}\:{should} \\ $$$${tangent}\:{both}\:{y}−{axis}\:{and}\:{x}−{axis}. \\ $$
Commented by ajfour last updated on 30/Apr/25
Yes! but if we wisely take (r−x)^2 +y^2 =r^2 y=x^2 +(k^2 −r)=x^2 +k_(new) ^2 and then solve, we should then get the radii r_(old) =r_(new) .
$${Yes}!\:{but}\:{if}\:{we}\:{wisely}\:{take} \\ $$$$\left({r}−{x}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${y}={x}^{\mathrm{2}} +\left({k}^{\mathrm{2}} −{r}\right)={x}^{\mathrm{2}} +{k}_{{new}} ^{\mathrm{2}} \\ $$$${and}\:{then}\:{solve},\:{we}\:{should}\:{then}\:{get}\:{the} \\ $$$${radii}\:{r}_{{old}} ={r}_{{new}} . \\ $$
Commented by mr W last updated on 30/Apr/25
k^2 −r is no constant more. this makes the equations hard to solve.
$${k}^{\mathrm{2}} −{r}\:{is}\:{no}\:{constant}\:{more}.\:{this} \\ $$$${makes}\:{the}\:{equations}\:{hard}\:{to}\:{solve}. \\ $$
Commented by mr W last updated on 30/Apr/25
i mean in this case the app is also unable to find a formula for r in terms of k.
$${i}\:{mean}\:{in}\:{this}\:{case}\:{the}\:{app}\:{is}\:{also} \\ $$$${unable}\:{to}\:{find}\:{a}\:{formula}\:{for}\: \\ $$$${r}\:{in}\:{terms}\:{of}\:{k}. \\ $$
Commented by ajfour last updated on 30/Apr/25
yes sir! unique question∪
$${yes}\:{sir}!\:{unique}\:{question}\cup \\ $$

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