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Question-62839

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Question Number 62839 by ajfour last updated on 25/Jun/19
Commented by ajfour last updated on 25/Jun/19
let θ be angle between b and extended c . (c+bcos θ)^2 +b^2 sin^2 θ=r^2 (c+asin θ)^2 +a^2 cos^2 θ=r^2 ⇒ b^2 +c^2 +2bccos θ=r^2 a^2 +c^2 +2acsin θ=r^2 ⇒ 2c(asin θ−bcos θ)=b^2 −a^2 2c(√(a^2 +b^2 ))sin (θ−tan^(−1) (b/a))=b^2 −a^2 ⇒ θ=α+sin^(−1) (p/h) = α+β .....(I) but on adding 2r^2 = a^2 +b^2 +2c^2 +2c(√(a^2 +b^2 ))sin (θ+tan^(−1) (b/a)) ⇒ r^2 =c^2 +((a^2 +b^2 )/2)+(h/2)sin (2α+β) = c^2 +((a^2 +b^2 )/2)+(h/2){((2ab)/(a^2 +b^2 ))×(√(1−(p^2 /h^2 )))+(((2a^2 )/(a^2 +b^2 ))−1)(p/h)} r^2 =c^2 +((a^2 +b^2 )/2)+((ab(√(h^2 −p^2 )))/(a^2 +b^2 ))+(((a^2 −b^2 )p)/(2(a^2 +b^2 ))) r^2 =c^2 +((a^2 +b^2 )/2)−(((b^2 −a^2 )^2 )/(2(a^2 +b^2 )))+((ab)/((a^2 +b^2 )))(√(4c^2 (a^2 +b^2 )−(b^2 −a^2 )^2 )) ⇒ r^2 =c^2 +((2a^2 b^2 )/(a^2 +b^2 ))+((ab)/((a^2 +b^2 )))(√(4c^2 (a^2 +b^2 )−(b^2 −a^2 )^2 )) r=(√(c^2 +((2a^2 b^2 )/(a^2 +b^2 ))+((ab)/((a^2 +b^2 )))(√(4c^2 (a^2 +b^2 )−(b^2 −a^2 )^2 )))) .
$${let}\:\theta\:{be}\:{angle}\:{between}\:{b}\:{and}\: \\ $$$${extended}\:{c}\:. \\ $$$$\left({c}+{b}\mathrm{cos}\:\theta\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta={r}^{\mathrm{2}} \\ $$$$\left({c}+{a}\mathrm{sin}\:\theta\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta={r}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\:\:\:\:{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{bc}\mathrm{cos}\:\theta={r}^{\mathrm{2}} \\ $$$$\:\:\:\:{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}\mathrm{sin}\:\theta={r}^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{2}{c}\left({a}\mathrm{sin}\:\theta−{b}\mathrm{cos}\:\theta\right)={b}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$$\mathrm{2}{c}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\mathrm{sin}\:\left(\theta−\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{a}}\right)={b}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\theta=\alpha+\mathrm{sin}^{−\mathrm{1}} \frac{{p}}{{h}}\:=\:\alpha+\beta\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..\left({I}\right) \\ $$$${but}\:{on}\:{adding} \\ $$$$\:\mathrm{2}{r}^{\mathrm{2}} =\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} +\mathrm{2}{c}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\mathrm{sin}\:\left(\theta+\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{a}}\right) \\ $$$$\Rightarrow\:{r}^{\mathrm{2}} ={c}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}+\frac{{h}}{\mathrm{2}}\mathrm{sin}\:\left(\mathrm{2}\alpha+\beta\right) \\ $$$$\:\:\:\:=\:{c}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}+\frac{{h}}{\mathrm{2}}\left\{\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }×\sqrt{\mathrm{1}−\frac{{p}^{\mathrm{2}} }{{h}^{\mathrm{2}} }}+\left(\frac{\mathrm{2}{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−\mathrm{1}\right)\frac{{p}}{{h}}\right\} \\ $$$${r}^{\mathrm{2}} ={c}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}+\frac{{ab}\sqrt{{h}^{\mathrm{2}} −{p}^{\mathrm{2}} }}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){p}}{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} \\ $$$${r}^{\mathrm{2}} ={c}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}−\frac{\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}+\frac{{ab}}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}\sqrt{\mathrm{4}{c}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:{r}^{\mathrm{2}} ={c}^{\mathrm{2}} +\frac{\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\frac{{ab}}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}\sqrt{\mathrm{4}{c}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${r}=\sqrt{{c}^{\mathrm{2}} +\frac{\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\frac{{ab}}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}\sqrt{\mathrm{4}{c}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} }}\:\:. \\ $$
Commented by ajfour last updated on 25/Jun/19
Find radius of circle in terms of a,b,c . C is center of circle, and segments of lengths a and b are perpendicular to each other.
$${Find}\:{radius}\:{of}\:{circle}\:{in}\:{terms}\:{of} \\ $$$${a},{b},{c}\:.\:{C}\:{is}\:{center}\:{of}\:{circle},\:{and} \\ $$$${segments}\:{of}\:{lengths}\:{a}\:{and}\:{b}\:{are} \\ $$$${perpendicular}\:{to}\:{each}\:{other}. \\ $$
Commented by behi83417@gmail.com last updated on 25/Jun/19
great question,sir Ajfour! waitng for solution(s)..........
$$\mathrm{great}\:\mathrm{question},\mathrm{sir}\:\mathrm{Ajfour}! \\ $$$$\mathrm{waitng}\:\mathrm{for}\:\mathrm{solution}\left(\mathrm{s}\right)………. \\ $$
Answered by mr W last updated on 25/Jun/19
r=radius of circle α=angle between a and c β=angle between b and c α+β+(π/2)=2π ⇒α=2π−((π/2)+β) ⇒cos α=−sin β ⇒cos^2 α+cos^2 β=1 cos α=((a^2 +c^2 −r^2 )/(2ac)) cos β=((b^2 +c^2 −r^2 )/(2bc)) ⇒(((a^2 +c^2 −r^2 )/(2ac)))^2 +(((b^2 +c^2 −r^2 )/(2bc)))^2 =1 b^2 [a^4 +2a^2 c^2 +c^4 −2(a^2 +c^2 )r^2 +r^4 ]+a^2 [b^4 +2b^2 c^2 +c^4 −2(b^2 +c^2 )r^2 +r^4 ]=4a^2 b^2 c^2 (a^2 +b^2 )r^4 −2[2a^2 b^2 +(a^2 +b^2 )c^2 ]r^2 +(a^2 +b^2 )(a^2 b^2 +c^4 )=0 ⇒r^2 =((2a^2 b^2 +(a^2 +b^2 )c^2 +(√([2a^2 b^2 +(a^2 +b^2 )c^2 ]^2 −(a^2 +b^2 )^2 (a^2 b^2 +c^4 ))))/(a^2 +b^2 )) ⇒r^2 =((2a^2 b^2 +(a^2 +b^2 )c^2 +ab(√(4(a^2 +b^2 )c^2 −(a^2 −b^2 )^2 )))/(a^2 +b^2 )) ⇒r=(√((2a^2 b^2 +(a^2 +b^2 )c^2 +ab(√(4(a^2 +b^2 )c^2 −(a^2 −b^2 )^2 )))/(a^2 +b^2 )))
$${r}={radius}\:{of}\:{circle} \\ $$$$\alpha={angle}\:{between}\:{a}\:{and}\:{c} \\ $$$$\beta={angle}\:{between}\:{b}\:{and}\:{c} \\ $$$$\alpha+\beta+\frac{\pi}{\mathrm{2}}=\mathrm{2}\pi \\ $$$$\Rightarrow\alpha=\mathrm{2}\pi−\left(\frac{\pi}{\mathrm{2}}+\beta\right) \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=−\mathrm{sin}\:\beta \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{2}} \:\alpha+\mathrm{cos}^{\mathrm{2}} \:\beta=\mathrm{1} \\ $$$$\mathrm{cos}\:\alpha=\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{ac}} \\ $$$$\mathrm{cos}\:\beta=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$$\Rightarrow\left(\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{ac}}\right)^{\mathrm{2}} +\left(\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{bc}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${b}^{\mathrm{2}} \left[{a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{4}} −\mathrm{2}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right){r}^{\mathrm{2}} +{r}^{\mathrm{4}} \right]+{a}^{\mathrm{2}} \left[{b}^{\mathrm{4}} +\mathrm{2}{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{4}} −\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right){r}^{\mathrm{2}} +{r}^{\mathrm{4}} \right]=\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){r}^{\mathrm{4}} −\mathrm{2}\left[\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){c}^{\mathrm{2}} \right]{r}^{\mathrm{2}} +\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{c}^{\mathrm{4}} \right)=\mathrm{0} \\ $$$$\Rightarrow{r}^{\mathrm{2}} =\frac{\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){c}^{\mathrm{2}} +\sqrt{\left[\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){c}^{\mathrm{2}} \right]^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} \left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{c}^{\mathrm{4}} \right)}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{r}^{\mathrm{2}} =\frac{\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){c}^{\mathrm{2}} +{ab}\sqrt{\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){c}^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} }}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{r}=\sqrt{\frac{\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){c}^{\mathrm{2}} +{ab}\sqrt{\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){c}^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} }}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$
Commented by ajfour last updated on 25/Jun/19
Too good Sir. Thanks for confirming.
$${Too}\:{good}\:{Sir}.\:{Thanks}\:{for}\:{confirming}. \\ $$

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