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Question-38092

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Question Number 38092 by ajfour last updated on 21/Jun/18
Commented by ajfour last updated on 22/Jun/18
The circle touches x=0 , y=0 , and the ellipse (x^2 /a^2 )+(y^2 /b^2 )=1 , in the manner shown; find its radius R in terms of a and b.
$${The}\:{circle}\:{touches}\:{x}=\mathrm{0}\:,\:{y}=\mathrm{0}\:, \\ $$$${and}\:{the}\:{ellipse}\:\:\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:,\:{in}\:{the} \\ $$$${manner}\:{shown};\:{find}\:{its}\:{radius}\:{R} \\ $$$${in}\:{terms}\:{of}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}. \\ $$$$\:\: \\ $$
Commented by MJS last updated on 22/Jun/18
there are 2 circles: the one you show and a small one within the ellipse. I′m afraid this leads to polynomes of 4^(th) or even 6^(th) degree...
$$\mathrm{there}\:\mathrm{are}\:\mathrm{2}\:\mathrm{circles}:\:\mathrm{the}\:\mathrm{one}\:\mathrm{you}\:\mathrm{show}\:\mathrm{and}\:\mathrm{a} \\ $$$$\mathrm{small}\:\mathrm{one}\:\mathrm{within}\:\mathrm{the}\:\mathrm{ellipse}. \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{afraid}\:\mathrm{this}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{polynomes}\:\mathrm{of}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{or} \\ $$$$\mathrm{even}\:\mathrm{6}^{\mathrm{th}} \:\mathrm{degree}… \\ $$
Commented by MJS last updated on 22/Jun/18
after not succeeding from the front I tried it from the rear: start with the common tangent in the intersection point P t: y=kx+d; k<0; d>0 P∈t: P= ((p),((kp+d)) ) n⊥t; P∈n n: y=−(1/k)x+d+(((k^2 +1)p)/k) the centers of the circles lie on m: y=x m∩n x=−(1/k)x+d+(((k^2 +1)p)/k) ⇒ x=((dk+(k^2 +1)p)/(k+1)) C= ((((dk+(k^2 +1)p)/(k+1))),(((dk+(k^2 +1)p)/(k+1))) ) ∣CP∣=r=x (((dk+(k^2 +1)p)/(k+1))−p)^2 +(((dk+(k^2 +1)p)/(k+1))−kp−d)^2 −(((dk+(k^2 +1)p)/(k+1)))^2 =0 ⇒ { ((p_1 =(d/(2k))(−1−((k+1)/( (√(k^2 +1))))))),((p_2 =(d/(2k))(−1+((k+1)/( (√(k^2 +1))))))) :} ⇒ { ((P_1 = ((((d/(2k))(−1−((k+1)/( (√(k^2 +1))))))),(((d/2)(1−((k+1)/( (√(k^2 +1))))))) ))),((P_2 = ((((d/(2k))(−1+((k+1)/( (√(k^2 +1))))))),(((d/2)(1+((k+1)/( (√(k^2 +1))))))) ))) :} ⇒ { ((C_1 = ((r_1 ),(r_1 ) ); r_1 =(d/(2k))(k−1−(√(k^2 +1))))),((C_2 = ((r_2 ),(r_2 ) ); r_2 =(d/(2k))(k−1+(√(k^2 +1))))) :} now we can find 2 ellipses for given k and d t_P ^(ell) =−((bp)/(a(√(a^2 −p^2 ))))x+((ab)/( (√(a^2 −p^2 )))) = kx+d ⇒ k=−((bp)/(a(√(a^2 −p^2 )))); d=((ab)/( (√(a^2 −p^2 )))) ⇒ a=(√(dp/(−k))); b=(√(d(d+kp)))
$$\mathrm{after}\:\mathrm{not}\:\mathrm{succeeding}\:\mathrm{from}\:\mathrm{the}\:\mathrm{front}\:\mathrm{I}\:\mathrm{tried} \\ $$$$\mathrm{it}\:\mathrm{from}\:\mathrm{the}\:\mathrm{rear}: \\ $$$$ \\ $$$$\mathrm{start}\:\mathrm{with}\:\mathrm{the}\:\mathrm{common}\:\mathrm{tangent}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{intersection}\:\mathrm{point}\:{P} \\ $$$${t}:\:{y}={kx}+{d};\:{k}<\mathrm{0};\:{d}>\mathrm{0} \\ $$$${P}\in{t}:\:{P}=\begin{pmatrix}{{p}}\\{{kp}+{d}}\end{pmatrix} \\ $$$${n}\bot{t};\:{P}\in{n} \\ $$$${n}:\:{y}=−\frac{\mathrm{1}}{{k}}{x}+{d}+\frac{\left({k}^{\mathrm{2}} +\mathrm{1}\right){p}}{{k}} \\ $$$$\mathrm{the}\:\mathrm{centers}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circles}\:\mathrm{lie}\:\mathrm{on} \\ $$$${m}:\:{y}={x} \\ $$$${m}\cap{n} \\ $$$${x}=−\frac{\mathrm{1}}{{k}}{x}+{d}+\frac{\left({k}^{\mathrm{2}} +\mathrm{1}\right){p}}{{k}}\:\Rightarrow\:{x}=\frac{{dk}+\left({k}^{\mathrm{2}} +\mathrm{1}\right){p}}{{k}+\mathrm{1}} \\ $$$${C}=\begin{pmatrix}{\frac{{dk}+\left({k}^{\mathrm{2}} +\mathrm{1}\right){p}}{{k}+\mathrm{1}}}\\{\frac{{dk}+\left({k}^{\mathrm{2}} +\mathrm{1}\right){p}}{{k}+\mathrm{1}}}\end{pmatrix} \\ $$$$\mid{CP}\mid={r}={x} \\ $$$$\left(\frac{{dk}+\left({k}^{\mathrm{2}} +\mathrm{1}\right){p}}{{k}+\mathrm{1}}−{p}\right)^{\mathrm{2}} +\left(\frac{{dk}+\left({k}^{\mathrm{2}} +\mathrm{1}\right){p}}{{k}+\mathrm{1}}−{kp}−{d}\right)^{\mathrm{2}} −\left(\frac{{dk}+\left({k}^{\mathrm{2}} +\mathrm{1}\right){p}}{{k}+\mathrm{1}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\begin{cases}{{p}_{\mathrm{1}} =\frac{{d}}{\mathrm{2}{k}}\left(−\mathrm{1}−\frac{{k}+\mathrm{1}}{\:\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}}\right)}\\{{p}_{\mathrm{2}} =\frac{{d}}{\mathrm{2}{k}}\left(−\mathrm{1}+\frac{{k}+\mathrm{1}}{\:\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}}\right)}\end{cases} \\ $$$$\Rightarrow\:\begin{cases}{{P}_{\mathrm{1}} =\begin{pmatrix}{\frac{{d}}{\mathrm{2}{k}}\left(−\mathrm{1}−\frac{{k}+\mathrm{1}}{\:\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}}\right)}\\{\frac{{d}}{\mathrm{2}}\left(\mathrm{1}−\frac{{k}+\mathrm{1}}{\:\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}}\right)}\end{pmatrix}}\\{{P}_{\mathrm{2}} =\begin{pmatrix}{\frac{{d}}{\mathrm{2}{k}}\left(−\mathrm{1}+\frac{{k}+\mathrm{1}}{\:\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}}\right)}\\{\frac{{d}}{\mathrm{2}}\left(\mathrm{1}+\frac{{k}+\mathrm{1}}{\:\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}}\right)}\end{pmatrix}}\end{cases} \\ $$$$\Rightarrow\:\begin{cases}{{C}_{\mathrm{1}} =\begin{pmatrix}{{r}_{\mathrm{1}} }\\{{r}_{\mathrm{1}} }\end{pmatrix};\:{r}_{\mathrm{1}} =\frac{{d}}{\mathrm{2}{k}}\left({k}−\mathrm{1}−\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}\right)}\\{{C}_{\mathrm{2}} =\begin{pmatrix}{{r}_{\mathrm{2}} }\\{{r}_{\mathrm{2}} }\end{pmatrix};\:{r}_{\mathrm{2}} =\frac{{d}}{\mathrm{2}{k}}\left({k}−\mathrm{1}+\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}\right)}\end{cases} \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{can}\:\mathrm{find}\:\mathrm{2}\:\mathrm{ellipses}\:\mathrm{for}\:\mathrm{given}\:{k}\:\mathrm{and}\:{d} \\ $$$${t}_{{P}} ^{\mathrm{ell}} =−\frac{{bp}}{{a}\sqrt{{a}^{\mathrm{2}} −{p}^{\mathrm{2}} }}{x}+\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} −{p}^{\mathrm{2}} }}\:=\:{kx}+{d} \\ $$$$\Rightarrow\:{k}=−\frac{{bp}}{{a}\sqrt{{a}^{\mathrm{2}} −{p}^{\mathrm{2}} }};\:{d}=\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} −{p}^{\mathrm{2}} }} \\ $$$$\Rightarrow\:{a}=\sqrt{\frac{{dp}}{−{k}}};\:{b}=\sqrt{{d}\left({d}+{kp}\right)} \\ $$
Commented by ajfour last updated on 23/Jun/18
Excellent! Sir. aint your expressions for p_1 , p_2 a bit simpler; i dont get the same..
$${Excellent}!\:{Sir}. \\ $$$${aint}\:{your}\:{expressions}\:{for}\:{p}_{\mathrm{1}} ,\:{p}_{\mathrm{2}} \: \\ $$$${a}\:{bit}\:{simpler};\:{i}\:{dont}\:{get}\:{the}\:{same}.. \\ $$
Commented by MJS last updated on 25/Jun/18
(((dk+(k^2 +1)p)/(k+1))−p)^2 +(((dk+(k^2 +1)p)/(k+1))−kp−d)^2 −(((dk+(k^2 +1)p)/(k+1)))^2 =0 (((dk+k(k−1)p)^2 +(−d−(k−1)p)^2 −(dk+(k^2 +1)p)^2 )/((k+1)^2 ))=0 k^2 (k−1)^2 p^2 +2dk^2 (k−1)p+d^2 k^2 + +(k−1)^2 p^2 +2d(k−1)p+d^2 − −((k^2 +1)^2 p^2 +2dk(k^2 +1)p+d^2 k^2 )=0 −2k(k^2 +1)p^2 −2d(k^2 +1)p+d^2 =0 p^2 +(d/k)p−(d^2 /(2k(k^2 +1)))=0 p=−(d/(2k))±(√((d^2 /(4k^2 ))+(d^2 /(2k(k^2 +1)))))= =−(d/(2k))±(√((d^2 (k^2 +1)+2d^2 k)/(4k^2 (k^2 +1))))= =−(d/(2k))±(d/(2k))(√((k^2 +1+2k)/(k^2 +1)))= =(d/(2k))(−1±((k+1)/( (√(k^2 +1)))))
$$\left(\frac{{dk}+\left({k}^{\mathrm{2}} +\mathrm{1}\right){p}}{{k}+\mathrm{1}}−{p}\right)^{\mathrm{2}} +\left(\frac{{dk}+\left({k}^{\mathrm{2}} +\mathrm{1}\right){p}}{{k}+\mathrm{1}}−{kp}−{d}\right)^{\mathrm{2}} −\left(\frac{{dk}+\left({k}^{\mathrm{2}} +\mathrm{1}\right){p}}{{k}+\mathrm{1}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\frac{\left({dk}+{k}\left({k}−\mathrm{1}\right){p}\right)^{\mathrm{2}} +\left(−{d}−\left({k}−\mathrm{1}\right){p}\right)^{\mathrm{2}} −\left({dk}+\left({k}^{\mathrm{2}} +\mathrm{1}\right){p}\right)^{\mathrm{2}} }{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$${k}^{\mathrm{2}} \left({k}−\mathrm{1}\right)^{\mathrm{2}} {p}^{\mathrm{2}} +\mathrm{2}{dk}^{\mathrm{2}} \left({k}−\mathrm{1}\right){p}+{d}^{\mathrm{2}} {k}^{\mathrm{2}} + \\ $$$$\:\:\:\:\:+\left({k}−\mathrm{1}\right)^{\mathrm{2}} {p}^{\mathrm{2}} +\mathrm{2}{d}\left({k}−\mathrm{1}\right){p}+{d}^{\mathrm{2}} − \\ $$$$\:\:\:\:\:\:\:\:\:\:−\left(\left({k}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} {p}^{\mathrm{2}} +\mathrm{2}{dk}\left({k}^{\mathrm{2}} +\mathrm{1}\right){p}+{d}^{\mathrm{2}} {k}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$ \\ $$$$−\mathrm{2}{k}\left({k}^{\mathrm{2}} +\mathrm{1}\right){p}^{\mathrm{2}} −\mathrm{2}{d}\left({k}^{\mathrm{2}} +\mathrm{1}\right){p}+{d}^{\mathrm{2}} =\mathrm{0} \\ $$$${p}^{\mathrm{2}} +\frac{{d}}{{k}}{p}−\frac{{d}^{\mathrm{2}} }{\mathrm{2}{k}\left({k}^{\mathrm{2}} +\mathrm{1}\right)}=\mathrm{0} \\ $$$${p}=−\frac{{d}}{\mathrm{2}{k}}\pm\sqrt{\frac{{d}^{\mathrm{2}} }{\mathrm{4}{k}^{\mathrm{2}} }+\frac{{d}^{\mathrm{2}} }{\mathrm{2}{k}\left({k}^{\mathrm{2}} +\mathrm{1}\right)}}= \\ $$$$=−\frac{{d}}{\mathrm{2}{k}}\pm\sqrt{\frac{{d}^{\mathrm{2}} \left({k}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{2}{d}^{\mathrm{2}} {k}}{\mathrm{4}{k}^{\mathrm{2}} \left({k}^{\mathrm{2}} +\mathrm{1}\right)}}= \\ $$$$=−\frac{{d}}{\mathrm{2}{k}}\pm\frac{{d}}{\mathrm{2}{k}}\sqrt{\frac{{k}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{k}}{{k}^{\mathrm{2}} +\mathrm{1}}}= \\ $$$$=\frac{{d}}{\mathrm{2}{k}}\left(−\mathrm{1}\pm\frac{{k}+\mathrm{1}}{\:\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}}\right) \\ $$

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