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Question-150363

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Question Number 150363 by ajfour last updated on 11/Aug/21
Commented by ajfour last updated on 11/Aug/21
If there is equal length in between consecutive black dots, find equation of such line.
$${If}\:{there}\:{is}\:{equal}\:{length}\:{in} \\ $$$${between}\:{consecutive}\:{black} \\ $$$${dots},\:{find}\:{equation}\:{of}\:{such} \\ $$$${line}. \\ $$
Commented by mr W last updated on 11/Aug/21
it′s possible and the right most point lies always on the x−axis.
$${it}'{s}\:{possible}\:{and}\:{the}\:{right}\:{most}\:{point} \\ $$$${lies}\:{always}\:{on}\:{the}\:{x}−{axis}. \\ $$
Commented by mr W last updated on 11/Aug/21
Commented by mr W last updated on 11/Aug/21
Commented by mr W last updated on 11/Aug/21
Answered by mr W last updated on 12/Aug/21
say the line is y=mx+c intersection with circle 1: x^2 +y^2 =b^2 x^2 +(mx+c)^2 =b^2 (1+m^2 )x^2 +2mcx+c^2 −b^2 =0 (x_3 −x_1 )^2 =(x_3 +x_1 )^2 −4x_3 x_1 (x_3 −x_1 )^2 =(−((2mc)/(1+m^2 )))^2 −4×((c^2 −b^2 )/(1+m^2 )) intersection with circle 2: (x−a)^2 +y^2 =a^2 (x−a)^2 +(mx+c)^2 =a^2 (1+m^2 )x^2 +2(mc−a)x+c^2 =0 (x_4 −x_2 )^2 =(x_4 +x_2 )^2 −4x_4 x_2 (x_4 −x_2 )^2 =(−((2(mc−a))/(1+m^2 )))^2 −4×(c^2 /(1+m^2 )) x_4 −x_2 =x_3 −x_1 ((m^2 c^2 )/((1+m^2 )^2 ))−((c^2 −b^2 )/(1+m^2 ))=(((mc−a)^2 )/((1+m^2 )^2 ))−(c^2 /(1+m^2 )) (2mc−a)a+(1+m^2 )b^2 =0 ⇒(1+m^2 )b^2 +2mac−a^2 =0 ...(i) ((x_4 +x_2 )/2)−((x_3 +x_1 )/2)=((x_3 −x_1 )/2) −((2(mc−a))/(1+m^2 ))+((2mc)/(1+m^2 ))=2(√((((mc)/(1+m^2 )))^2 −((c^2 −b^2 )/(1+m^2 )))) a=(√(m^2 c^2 −(1+m^2 )(c^2 −b^2 ))) ⇒c^2 =(1+m^2 )b^2 −a^2 ...(ii) c^2 +2mac=0 ⇒c=−2ma m^2 (4a^2 −b^2 )=b^2 −a^2 ⇒m=(√((b^2 −a^2 )/(4a^2 −b^2 ))) ⇒c=−2a(√((b^2 −a^2 )/(4a^2 −b^2 ))) with a≤b<2a ⇒y=(x−2a)(√((b^2 −a^2 )/(4a^2 +b^2 )))
$${say}\:{the}\:{line}\:{is}\:{y}={mx}+{c} \\ $$$${intersection}\:{with}\:{circle}\:\mathrm{1}: \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\left({mx}+{c}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+{m}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\mathrm{2}{mcx}+{c}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}_{\mathrm{3}} −{x}_{\mathrm{1}} \right)^{\mathrm{2}} =\left({x}_{\mathrm{3}} +{x}_{\mathrm{1}} \right)^{\mathrm{2}} −\mathrm{4}{x}_{\mathrm{3}} {x}_{\mathrm{1}} \\ $$$$\left({x}_{\mathrm{3}} −{x}_{\mathrm{1}} \right)^{\mathrm{2}} =\left(−\frac{\mathrm{2}{mc}}{\mathrm{1}+{m}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{4}×\frac{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{1}+{m}^{\mathrm{2}} } \\ $$$${intersection}\:{with}\:{circle}\:\mathrm{2}: \\ $$$$\left({x}−{a}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\left({x}−{a}\right)^{\mathrm{2}} +\left({mx}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+{m}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\mathrm{2}\left({mc}−{a}\right){x}+{c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}_{\mathrm{4}} −{x}_{\mathrm{2}} \right)^{\mathrm{2}} =\left({x}_{\mathrm{4}} +{x}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{x}_{\mathrm{4}} {x}_{\mathrm{2}} \\ $$$$\left({x}_{\mathrm{4}} −{x}_{\mathrm{2}} \right)^{\mathrm{2}} =\left(−\frac{\mathrm{2}\left({mc}−{a}\right)}{\mathrm{1}+{m}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{4}×\frac{{c}^{\mathrm{2}} }{\mathrm{1}+{m}^{\mathrm{2}} } \\ $$$${x}_{\mathrm{4}} −{x}_{\mathrm{2}} ={x}_{\mathrm{3}} −{x}_{\mathrm{1}} \\ $$$$\frac{{m}^{\mathrm{2}} {c}^{\mathrm{2}} }{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\mathrm{2}} }−\frac{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{1}+{m}^{\mathrm{2}} }=\frac{\left({mc}−{a}\right)^{\mathrm{2}} }{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\mathrm{2}} }−\frac{{c}^{\mathrm{2}} }{\mathrm{1}+{m}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}{mc}−{a}\right){a}+\left(\mathrm{1}+{m}^{\mathrm{2}} \right){b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}+{m}^{\mathrm{2}} \right){b}^{\mathrm{2}} +\mathrm{2}{mac}−{a}^{\mathrm{2}} =\mathrm{0}\:\:\:…\left({i}\right) \\ $$$$\frac{{x}_{\mathrm{4}} +{x}_{\mathrm{2}} }{\mathrm{2}}−\frac{{x}_{\mathrm{3}} +{x}_{\mathrm{1}} }{\mathrm{2}}=\frac{{x}_{\mathrm{3}} −{x}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$−\frac{\mathrm{2}\left({mc}−{a}\right)}{\mathrm{1}+{m}^{\mathrm{2}} }+\frac{\mathrm{2}{mc}}{\mathrm{1}+{m}^{\mathrm{2}} }=\mathrm{2}\sqrt{\left(\frac{{mc}}{\mathrm{1}+{m}^{\mathrm{2}} }\right)^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{1}+{m}^{\mathrm{2}} }} \\ $$$${a}=\sqrt{{m}^{\mathrm{2}} {c}^{\mathrm{2}} −\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow{c}^{\mathrm{2}} =\left(\mathrm{1}+{m}^{\mathrm{2}} \right){b}^{\mathrm{2}} −{a}^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$${c}^{\mathrm{2}} +\mathrm{2}{mac}=\mathrm{0}\: \\ $$$$\Rightarrow{c}=−\mathrm{2}{ma} \\ $$$${m}^{\mathrm{2}} \left(\mathrm{4}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)={b}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$$\Rightarrow{m}=\sqrt{\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }} \\ $$$$\Rightarrow{c}=−\mathrm{2}{a}\sqrt{\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }} \\ $$$${with}\:{a}\leqslant{b}<\mathrm{2}{a} \\ $$$$\Rightarrow{y}=\left({x}−\mathrm{2}{a}\right)\sqrt{\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$
Commented by ajfour last updated on 12/Aug/21
Thank you sir, i followed it all.
Answered by ajfour last updated on 12/Aug/21
Commented by ajfour last updated on 12/Aug/21
For a=4, b=5 such a vertical line is x=(9/2)−((√(274))/4) ≈ 0.36176 (green vertical)
$${For}\:{a}=\mathrm{4},\:{b}=\mathrm{5} \\ $$$${such}\:{a}\:{vertical}\:{line}\:{is} \\ $$$$\:\:{x}=\frac{\mathrm{9}}{\mathrm{2}}−\frac{\sqrt{\mathrm{274}}}{\mathrm{4}}\:\:\approx\:\mathrm{0}.\mathrm{36176} \\ $$$$\:\:\left({green}\:{vertical}\right) \\ $$
Commented by ajfour last updated on 12/Aug/21
Commented by ajfour last updated on 12/Aug/21
b^2 =q^2 +4p^2 a^2 =q^2 +p^2 ⇒ p^2 =((b^2 −a^2 )/3) ; q^2 =((4a^2 −b^2 )/3) tan θ=m=(p/q)=(√((b^2 −a^2 )/(4a^2 −b^2 ))) Eq. of line_(−) : xsin θ−ycos θ=2p x(√(b^2 −a^2 ))−y(√(4a^2 −b^2 )) =((2(√(b^2 −a^2 )))/( (√3)))(a(√3)) ⇒ y=(√((b^2 −a^2 )/(4a^2 −b^2 ))) (x−2a) for e.g. if a=4, b=5 y=((3(x−8))/( (√(39)))) (yellow inclined) .............................................. but if line is vertical b^2 =(2p)^2 +(q+(q/2))^2 & a^2 =((q/2))^2 +(a−2p)^2 ⇒ 4p^2 =((4b^2 −9q^2 )/4) ⇒ 16p^2 =4b^2 −9×4{a^2 −(a−2p)^2 } ⇒ 4p^2 =b^2 −9(4ap−4p^2 ) ⇒ if p≠0, then 32p^2 −36ap+b^2 =0 p=((9a)/(16))±(√((81a^2 −8b^2 )/(256))) eq. of line x=2p x=((9a−(√(81a^2 −8b^2 )))/8) ∙ (+ sign rejected) Then for the blue vertical a^2 =(2p−a)^2 +(q+(q/2))^2 & b^2 =(2p)^2 +((q/2))^2 ⇒ 9b^2 =36p^2 +a^2 −(2p−a)^2 ⇒ 32p^2 +4ap−9b^2 =0 p=−(a/(16))+(√((a^2 +72b^2 )/(256))) for a=4, b=5 p=−(1/4)+((√(454))/8) blue line is x=2p=(((√(454))−2)/4) ≈ 4.8268
$${b}^{\mathrm{2}} ={q}^{\mathrm{2}} +\mathrm{4}{p}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} ={q}^{\mathrm{2}} +{p}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{p}^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{3}}\:\:;\:{q}^{\mathrm{2}} =\frac{\mathrm{4}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\mathrm{tan}\:\theta={m}=\frac{{p}}{{q}}=\sqrt{\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }} \\ $$$$\underset{−} {{Eq}.\:{of}\:{line}}:\:\:{x}\mathrm{sin}\:\theta−{y}\mathrm{cos}\:\theta=\mathrm{2}{p} \\ $$$${x}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }−{y}\sqrt{\mathrm{4}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\:\sqrt{\mathrm{3}}}\left({a}\sqrt{\mathrm{3}}\right) \\ $$$$\Rightarrow\:\:{y}=\sqrt{\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\:\left({x}−\mathrm{2}{a}\right) \\ $$$${for}\:{e}.{g}.\:\:{if}\:{a}=\mathrm{4},\:{b}=\mathrm{5} \\ $$$$\:\:\:\:{y}=\frac{\mathrm{3}\left({x}−\mathrm{8}\right)}{\:\sqrt{\mathrm{39}}}\:\:\:\left({yellow}\:{inclined}\right) \\ $$$$………………………………………. \\ $$$${but}\:{if}\:{line}\:{is}\:{vertical} \\ $$$${b}^{\mathrm{2}} =\left(\mathrm{2}{p}\right)^{\mathrm{2}} +\left({q}+\frac{{q}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\&\:\:\:{a}^{\mathrm{2}} =\left(\frac{{q}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({a}−\mathrm{2}{p}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{4}{p}^{\mathrm{2}} =\frac{\mathrm{4}{b}^{\mathrm{2}} −\mathrm{9}{q}^{\mathrm{2}} }{\mathrm{4}}\:\:\:\Rightarrow \\ $$$$\mathrm{16}{p}^{\mathrm{2}} =\mathrm{4}{b}^{\mathrm{2}} −\mathrm{9}×\mathrm{4}\left\{{a}^{\mathrm{2}} −\left({a}−\mathrm{2}{p}\right)^{\mathrm{2}} \right\} \\ $$$$\Rightarrow\:\:\mathrm{4}{p}^{\mathrm{2}} ={b}^{\mathrm{2}} −\mathrm{9}\left(\mathrm{4}{ap}−\mathrm{4}{p}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\:{if}\:\:{p}\neq\mathrm{0},\:\:{then} \\ $$$$\:\:\mathrm{32}{p}^{\mathrm{2}} −\mathrm{36}{ap}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${p}=\frac{\mathrm{9}{a}}{\mathrm{16}}\pm\sqrt{\frac{\mathrm{81}{a}^{\mathrm{2}} −\mathrm{8}{b}^{\mathrm{2}} }{\mathrm{256}}} \\ $$$${eq}.\:{of}\:{line}\:\:\:\:{x}=\mathrm{2}{p} \\ $$$$\:\:{x}=\frac{\mathrm{9}{a}−\sqrt{\mathrm{81}{a}^{\mathrm{2}} −\mathrm{8}{b}^{\mathrm{2}} }}{\mathrm{8}}\:\centerdot \\ $$$$\:\:\:\left(+\:{sign}\:{rejected}\right) \\ $$$${Then}\:{for}\:{the}\:{blue}\:{vertical} \\ $$$$\:\:{a}^{\mathrm{2}} =\left(\mathrm{2}{p}−{a}\right)^{\mathrm{2}} +\left({q}+\frac{{q}}{\mathrm{2}}\right)^{\mathrm{2}} \:\:\:\& \\ $$$$\:\:{b}^{\mathrm{2}} =\left(\mathrm{2}{p}\right)^{\mathrm{2}} +\left(\frac{{q}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{9}{b}^{\mathrm{2}} =\mathrm{36}{p}^{\mathrm{2}} +{a}^{\mathrm{2}} −\left(\mathrm{2}{p}−{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{32}{p}^{\mathrm{2}} +\mathrm{4}{ap}−\mathrm{9}{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:{p}=−\frac{{a}}{\mathrm{16}}+\sqrt{\frac{{a}^{\mathrm{2}} +\mathrm{72}{b}^{\mathrm{2}} }{\mathrm{256}}} \\ $$$${for}\:{a}=\mathrm{4},\:{b}=\mathrm{5} \\ $$$${p}=−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt{\mathrm{454}}}{\mathrm{8}} \\ $$$$\:\:{blue}\:{line}\:{is}\: \\ $$$$\:{x}=\mathrm{2}{p}=\frac{\sqrt{\mathrm{454}}−\mathrm{2}}{\mathrm{4}}\:\approx\:\mathrm{4}.\mathrm{8268} \\ $$$$ \\ $$
Commented by mr W last updated on 13/Aug/21
perfect! i didn′t think of the case with vertical line.
$${perfect}! \\ $$$${i}\:{didn}'{t}\:{think}\:{of}\:{the}\:{case}\:{with}\:{vertical} \\ $$$${line}. \\ $$

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