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Question Number 216060 by mr W last updated on 26/Jan/25
Commented by mr W last updated on 26/Jan/25
find radius of inscribed circle r=?
$${find}\:{radius}\:{of}\:{inscribed}\:{circle}\:{r}=? \\ $$
Commented by Tawa11 last updated on 18/May/25
Is the centre of inscribed circle one of the foci?
$$\mathrm{Is}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{inscribed}\:\mathrm{circle}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{foci}? \\ $$
Commented by mr W last updated on 18/May/25
maybe. it is only given that the inscribed circle tangents the y−axis.
$${maybe}. \\ $$$${it}\:{is}\:{only}\:{given}\:{that}\:{the}\:{inscribed} \\ $$$${circle}\:{tangents}\:{the}\:{y}−{axis}. \\ $$
Commented by mr W last updated on 19/May/25
what do think? can the center of the inscribed circle lie on the one focus of the ellipse?
$${what}\:{do}\:{think}?\:{can}\:{the}\:{center}\:{of} \\ $$$${the}\:{inscribed}\:{circle}\:{lie}\:{on}\:{the}\:{one} \\ $$$${focus}\:{of}\:{the}\:{ellipse}? \\ $$
Answered by mr W last updated on 26/Jan/25
say μ=(b/a), λ=(r/a) say P(a cos θ, b sin θ) tan φ=−(dx/dy)=((a sin θ)/(b cos θ))=((tan θ)/μ)=(m/μ) b sin θ=r sin φ ⇒μ sin θ=((λm)/( (√(m^2 +μ^2 )))) ...(i) a cos θ=r+r cos φ ⇒cos θ=λ(1+(μ/( (√(m^2 +μ^2 ))))) ...(ii) (i)/(ii): μ tan θ=(m/(μ+(√(m^2 +μ^2 )))) μ=(1/(μ+(√(m^2 +μ^2 )))) (√(m^2 +μ^2 ))=(1/μ)−μ ⇒m^2 =(1/μ^2 )−2 ≥0 ⇒μ≤(1/( (√2))) ⇒a≥(√2)b from (i): μ sin θ=((λm)/( (√(m^2 +μ^2 )))) ((μm)/( (√(1+m^2 ))))=((λm)/( (√(m^2 +μ^2 )))) ⇒λ=((μ (√(m^2 +μ^2 )))/( (√(1+m^2 ))))=((μ((1/μ)−μ))/( (√((1/μ^2 )−1))))=μ(√(1−μ^2 )) i.e. r=b(√(1−((b/a))^2 ))
$${say}\:\mu=\frac{{b}}{{a}},\:\lambda=\frac{{r}}{{a}} \\ $$$${say}\:{P}\left({a}\:\mathrm{cos}\:\theta,\:{b}\:\mathrm{sin}\:\theta\right) \\ $$$$\mathrm{tan}\:\phi=−\frac{{dx}}{{dy}}=\frac{{a}\:\mathrm{sin}\:\theta}{{b}\:\mathrm{cos}\:\theta}=\frac{\mathrm{tan}\:\theta}{\mu}=\frac{{m}}{\mu} \\ $$$${b}\:\mathrm{sin}\:\theta={r}\:\mathrm{sin}\:\phi \\ $$$$\Rightarrow\mu\:\mathrm{sin}\:\theta=\frac{\lambda{m}}{\:\sqrt{{m}^{\mathrm{2}} +\mu^{\mathrm{2}} }}\:\:\:…\left({i}\right) \\ $$$${a}\:\mathrm{cos}\:\theta={r}+{r}\:\mathrm{cos}\:\phi \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\lambda\left(\mathrm{1}+\frac{\mu}{\:\sqrt{{m}^{\mathrm{2}} +\mu^{\mathrm{2}} }}\right)\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\mu\:\mathrm{tan}\:\theta=\frac{{m}}{\mu+\sqrt{{m}^{\mathrm{2}} +\mu^{\mathrm{2}} }} \\ $$$$\mu=\frac{\mathrm{1}}{\mu+\sqrt{{m}^{\mathrm{2}} +\mu^{\mathrm{2}} }} \\ $$$$\sqrt{{m}^{\mathrm{2}} +\mu^{\mathrm{2}} }=\frac{\mathrm{1}}{\mu}−\mu \\ $$$$\Rightarrow{m}^{\mathrm{2}} =\frac{\mathrm{1}}{\mu^{\mathrm{2}} }−\mathrm{2}\:\geqslant\mathrm{0}\:\Rightarrow\mu\leqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\Rightarrow{a}\geqslant\sqrt{\mathrm{2}}{b} \\ $$$${from}\:\left({i}\right): \\ $$$$\mu\:\mathrm{sin}\:\theta=\frac{\lambda{m}}{\:\sqrt{{m}^{\mathrm{2}} +\mu^{\mathrm{2}} }} \\ $$$$\frac{\mu{m}}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}=\frac{\lambda{m}}{\:\sqrt{{m}^{\mathrm{2}} +\mu^{\mathrm{2}} }} \\ $$$$\Rightarrow\lambda=\frac{\mu\:\sqrt{{m}^{\mathrm{2}} +\mu^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}=\frac{\mu\left(\frac{\mathrm{1}}{\mu}−\mu\right)}{\:\sqrt{\frac{\mathrm{1}}{\mu^{\mathrm{2}} }−\mathrm{1}}}=\mu\sqrt{\mathrm{1}−\mu^{\mathrm{2}} } \\ $$$${i}.{e}.\:{r}={b}\sqrt{\mathrm{1}−\left(\frac{{b}}{{a}}\right)^{\mathrm{2}} } \\ $$
Commented by mr W last updated on 26/Jan/25
Commented by mr W last updated on 26/Jan/25
Commented by ajfour last updated on 27/Jan/25
If a≤(√2)b then such r=(a/2) And if a≥(√2)b r=(b/a)(√(a^2 −b^2 )) Isnt it this to summarize, Sir? Thanks.
$${If}\:{a}\leqslant\sqrt{\mathrm{2}}{b}\:\:\:{then}\:\:\:{such}\:\:{r}=\frac{{a}}{\mathrm{2}} \\ $$$${And}\:{if}\:\:{a}\geqslant\sqrt{\mathrm{2}}{b}\:\:\:\:\:{r}=\frac{{b}}{{a}}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$${Isnt}\:{it}\:{this}\:{to}\:{summarize},\:{Sir}?\:{Thanks}. \\ $$
Commented by mr W last updated on 27/Jan/25
if a=(√2)b then r=(a/2). if a<(√2)b then no inscribed circle possible.
$${if}\:{a}=\sqrt{\mathrm{2}}{b}\:{then}\:{r}=\frac{{a}}{\mathrm{2}}. \\ $$$${if}\:{a}<\sqrt{\mathrm{2}}{b}\:{then}\:{no}\:{inscribed}\:{circle}\: \\ $$$${possible}. \\ $$
Answered by aleks041103 last updated on 27/Jan/25
Commented by aleks041103 last updated on 27/Jan/25
The circle is tangent to the ellipse and therefore the radius is ⊥ to the tangent to the ellipse at this point. The tangent of an ellipse is ⊥ to the angle bisector of the angle between the segments connecting point the foci to the point. Thus we get the top schematic. Now: x+y=2a ⇒ y=2a−x (x/y)=((c+r)/(c−r)) xy−(c+r)(c−r)=r^2 ⇒ xy=c^2 =a^2 −b^2 c=a(√(1−((b/a))^2 )) ⇒x(2a−x)=a^2 −b^2 ⇒(x−a)^2 −b^2 =0 ⇒x=a+b;a−b but x>y ⇒ x=a+b, y=a−b ⇒(x/h)=((a+b)/(a−b))=((c+r)/(c−r)) ⇒r=((bc)/a) ⇒r=b(√(1−((b/a))^2 ))
$${The}\:{circle}\:{is}\:{tangent}\:{to}\:{the}\:{ellipse}\:{and}\:{therefore} \\ $$$${the}\:{radius}\:{is}\:\bot\:{to}\:{the}\:{tangent}\:{to}\:{the}\:{ellipse} \\ $$$${at}\:{this}\:{point}. \\ $$$${The}\:{tangent}\:{of}\:{an}\:{ellipse}\:{is}\:\bot\:{to}\:{the}\:{angle} \\ $$$${bisector}\:{of}\:{the}\:{angle}\:{between}\:{the}\:{segments} \\ $$$${connecting}\:{point}\:{the}\:{foci}\:{to}\:{the}\:{point}. \\ $$$${Thus}\:{we}\:{get}\:{the}\:{top}\:{schematic}. \\ $$$${Now}: \\ $$$${x}+{y}=\mathrm{2}{a}\:\Rightarrow\:{y}=\mathrm{2}{a}−{x} \\ $$$$\frac{{x}}{{y}}=\frac{{c}+{r}}{{c}−{r}} \\ $$$${xy}−\left({c}+{r}\right)\left({c}−{r}\right)={r}^{\mathrm{2}} \:\Rightarrow\:{xy}={c}^{\mathrm{2}} ={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$${c}={a}\sqrt{\mathrm{1}−\left(\frac{{b}}{{a}}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{x}\left(\mathrm{2}{a}−{x}\right)={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$$\Rightarrow\left({x}−{a}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{x}={a}+{b};{a}−{b} \\ $$$${but}\:{x}>{y}\:\Rightarrow\:{x}={a}+{b},\:{y}={a}−{b} \\ $$$$\Rightarrow\frac{{x}}{{h}}=\frac{{a}+{b}}{{a}−{b}}=\frac{{c}+{r}}{{c}−{r}} \\ $$$$\Rightarrow{r}=\frac{{bc}}{{a}} \\ $$$$\Rightarrow{r}={b}\sqrt{\mathrm{1}−\left(\frac{{b}}{{a}}\right)^{\mathrm{2}} } \\ $$
Commented by aleks041103 last updated on 27/Jan/25
The radius to the point P (original schematic) is perpendicular(⊥) to the tangent of the ellipse at P (since the tangent of the circle coinsides with the tangent to the ellipse).
$${The}\:{radius}\:{to}\:{the}\:{point}\:{P}\:\left({original}\:{schematic}\right) \\ $$$${is}\:{perpendicular}\left(\bot\right)\:{to}\:{the}\:{tangent}\:{of}\:{the} \\ $$$${ellipse}\:{at}\:{P}\:\left({since}\:{the}\:{tangent}\:{of}\:{the}\:{circle}\right. \\ $$$$\left.{coinsides}\:{with}\:{the}\:{tangent}\:{to}\:{the}\:{ellipse}\right). \\ $$
Commented by mr W last updated on 27/Jan/25
thanks for this nice solution!
$${thanks}\:{for}\:{this}\:{nice}\:{solution}! \\ $$

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