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Question Number 218526 by Spillover last updated on 11/Apr/25

Commented by mr W last updated on 11/Apr/25

≈0.8348

$$\approx\mathrm{0}.\mathrm{8348} \\ $$

Commented by Spillover last updated on 11/Apr/25

correct

$${correct} \\ $$

Answered by mr W last updated on 11/Apr/25

Commented by mr W last updated on 12/Apr/25

“a” in original question is replaced with 2s in following. say A(a,0), B(0,b) r=((ab)/(a+b+(√(a^2 +b^2 )))) (x/a)+(y/b)=1 or bx+ay=ab ← eqn. of AB s=((bs+as−ab)/( (√(a^2 +b^2 )))) (a+b−(√(a^2 +b^2 )))s=ab say α=(a/s), β=(b/s) α+β−(√(α^2 +β^2 ))=αβ β=((2(1−α))/(2−α)) ...(I) (x−a)^2 +y^2 =a^2 +b^2 x^2 −2ax+y^2 =b^2 ...(i) (x−s)^2 +(y−s)^2 =s^2 x^2 −2sx+y^2 −2sy=−s^2 ...(ii) (i)−(ii): 2(s−a)x+2sy=b^2 +s^2 ← eqn. of intersection line d=((2(s−a)s+2s^2 −b^2 −s^2 )/( 2(√((s−a)^2 +s^2 ))))=s−2r ((3s^2 −2as−b^2 )/( 2(√(2s^2 −2sa+a^2 ))))=s−((2ab)/(a+b+(√(a^2 +b^2 )))) ((3−2α−β^2 )/( 2(√(2−2α+α^2 ))))=1−((2αβ)/(α+β+(√(α^2 +β^2 )))) ...(II) from (I) and (II) we get α≈0.677 564 399 386, β≈0.487 639 020 704 ((2AB)/(“a”))=((AB)/s)=((√(a^2 +b^2 ))/s) =(√(α^2 +β^2 ))≈0.834 796 579 910 ✓

$$``{a}''\:{in}\:{original}\:{question}\:{is}\:{replaced} \\ $$$${with}\:\mathrm{2}{s}\:{in}\:{following}. \\ $$$${say}\:{A}\left({a},\mathrm{0}\right),\:{B}\left(\mathrm{0},{b}\right) \\ $$$${r}=\frac{{ab}}{{a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1}\:{or}\:{bx}+{ay}={ab}\:\:\:\leftarrow\:{eqn}.\:{of}\:{AB} \\ $$$${s}=\frac{{bs}+{as}−{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\left({a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right){s}={ab} \\ $$$${say}\:\alpha=\frac{{a}}{{s}},\:\beta=\frac{{b}}{{s}} \\ $$$$\alpha+\beta−\sqrt{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }=\alpha\beta \\ $$$$\beta=\frac{\mathrm{2}\left(\mathrm{1}−\alpha\right)}{\mathrm{2}−\alpha}\:\:\:...\left({I}\right) \\ $$$$ \\ $$$$\left({x}−{a}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{ax}+{y}^{\mathrm{2}} ={b}^{\mathrm{2}} \:\:\:...\left({i}\right) \\ $$$$\left({x}−{s}\right)^{\mathrm{2}} +\left({y}−{s}\right)^{\mathrm{2}} ={s}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{sx}+{y}^{\mathrm{2}} −\mathrm{2}{sy}=−{s}^{\mathrm{2}} \:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\mathrm{2}\left({s}−{a}\right){x}+\mathrm{2}{sy}={b}^{\mathrm{2}} +{s}^{\mathrm{2}} \:\:\:\leftarrow\:{eqn}.\:{of}\:{intersection}\:{line} \\ $$$${d}=\frac{\mathrm{2}\left({s}−{a}\right){s}+\mathrm{2}{s}^{\mathrm{2}} −{b}^{\mathrm{2}} −{s}^{\mathrm{2}} }{\:\mathrm{2}\sqrt{\left({s}−{a}\right)^{\mathrm{2}} +{s}^{\mathrm{2}} }}={s}−\mathrm{2}{r} \\ $$$$\frac{\mathrm{3}{s}^{\mathrm{2}} −\mathrm{2}{as}−{b}^{\mathrm{2}} }{\:\mathrm{2}\sqrt{\mathrm{2}{s}^{\mathrm{2}} −\mathrm{2}{sa}+{a}^{\mathrm{2}} }}={s}−\frac{\mathrm{2}{ab}}{{a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{3}−\mathrm{2}\alpha−\beta^{\mathrm{2}} }{\:\mathrm{2}\sqrt{\mathrm{2}−\mathrm{2}\alpha+\alpha^{\mathrm{2}} }}=\mathrm{1}−\frac{\mathrm{2}\alpha\beta}{\alpha+\beta+\sqrt{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }}\:\:\:...\left({II}\right) \\ $$$$ \\ $$$${from}\:\left({I}\right)\:{and}\:\left({II}\right)\:{we}\:{get} \\ $$$$\alpha\approx\mathrm{0}.\mathrm{677}\:\mathrm{564}\:\mathrm{399}\:\mathrm{386},\:\beta\approx\mathrm{0}.\mathrm{487}\:\mathrm{639}\:\mathrm{020}\:\mathrm{704} \\ $$$$\frac{\mathrm{2}{AB}}{``{a}''}=\frac{{AB}}{{s}}=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{s}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\sqrt{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }\approx\mathrm{0}.\mathrm{834}\:\mathrm{796}\:\mathrm{579}\:\mathrm{910}\:\checkmark \\ $$

Commented by mr W last updated on 11/Apr/25

Commented by Spillover last updated on 17/Apr/25

GOD bless you.very?nice solution

$${GOD}\:{bless}\:{you}.{very}?{nice} \\ $$$${solution} \\ $$

Answered by Spillover last updated on 11/Apr/25

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