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Question Number 48113 by ajfour last updated on 19/Nov/18

	Answered by MJS last updated on 20/Nov/18
	$circle 1 c_1 : (x+o)^2 +y^2 −o^2 =0 circle 2 c_2 : (x−p)^2 +y^2 −p^2 =0 let 0<o<p ellipse ell: b^2 (x−q)^2 +a^2 y^2 −a^2 b^2 =0 0<o<p ⇒ q>0 c_1 ∩ell={P_1 , P_2 }={ ((x),((−y)) ), ((x),((+y)) )} a^2 c_1 −ell: (a^2 −b^2 )x^2 +2(a^2 o+b^2 q)x+b^2 (a^2 −q^2 )=0 we need B^2 −4AC=0 ⇒ ⇒ a^2 =((b^2 (b^2 +2oq+q^2 ))/(b^2 −o^2 )) same for a^2 c_2 −ell ⇒ ⇒ a^2 =((b^2 (b^2 −2pq+q^2 ))/(b^2 −p^2 )) ((b^2 (b^2 +2oq+q^2 ))/(b^2 −o^2 ))=((b^2 (b^2 −2pq+q^2 ))/(b^2 −p^2 )) ∧ b≠0 ⇒ ⇒ b^2 =((q(2op+q(p−o)))/(2q−(p−o))) ⇒ a^2 =((2q^2 (2op+q(p−o)))/((p−o)(2q−(p−o)))) we now need the minimum of f(q)=ab f(q)=(((2op+q(p−o))(√(2q^3 )))/((2q−(p−o))(√(p−o)))) f′(q)=(√q)×((6q^2 (p−o)+(14op−5(o^2 +p^2 ))q−6op(p−o))/((2q−(p−o))(√(2(p−o))))) f′(q)=0 ∧ q≠0 ⇒ ⇒ q=((5o^2 −14op+5p^2 )/(12(p−o)))±(((o+p)(√(25o^2 −46op+25p^2 )))/(12(p−o))) trying with o=1 and p>1 ⇒ we need the solution with the + so we have a^2 =((2q^2 (2op+q(p−o)))/((p−o)(2q−(p−o)))) b^2 =((q(2op+q(p−o)))/(2q−(p−o))) q=((5o^2 −14op+5p^2 )/(12(p−o)))+(((o+p)(√(25o^2 −46op+25p^2 )))/(12(p−o)))$
	$circle 1$ $c_{1} : {(x + o)}^{2} + y^{2} - o^{2} = 0$ $circle 2$ $c_{2} : {(x - p)}^{2} + y^{2} - p^{2} = 0$ $let 0 < o < p$ $ellipse$ $e l l : b^{2} {(x - q)}^{2} + a^{2} y^{2} - a^{2} b^{2} = 0$ $0 < o < p \Rightarrow q > 0$ $c_{1} \cap e l l = {P_{1}, P_{2}} = {(\begin{matrix} x \\ - y \end{matrix}), (\begin{matrix} x \\ + y \end{matrix})}$ $a^{2} c_{1} - e l l :$ $(a^{2} - b^{2}) x^{2} + 2 (a^{2} o + b^{2} q) x + b^{2} (a^{2} - q^{2}) = 0$ $we need B^{2} - 4 A C = 0 \Rightarrow$ $\Rightarrow a^{2} = \frac{b^{2} (b^{2} + 2 o q + q^{2})}{b^{2} - o^{2}}$ $same for a^{2} c_{2} - e l l \Rightarrow$ $\Rightarrow a^{2} = \frac{b^{2} (b^{2} - 2 p q + q^{2})}{b^{2} - p^{2}}$ $\frac{b^{2} (b^{2} + 2 o q + q^{2})}{b^{2} - o^{2}} = \frac{b^{2} (b^{2} - 2 p q + q^{2})}{b^{2} - p^{2}} \land b \neq 0 \Rightarrow$ $\Rightarrow b^{2} = \frac{q (2 o p + q (p - o))}{2 q - (p - o)}$ $\Rightarrow a^{2} = \frac{2 q^{2} (2 o p + q (p - o))}{(p - o) (2 q - (p - o))}$ $we now need the minimum of f (q) = a b$ $f (q) = \frac{(2 o p + q (p - o)) \sqrt{2 q^{3}}}{(2 q - (p - o)) \sqrt{p - o}}$ $f^{'} (q) = \sqrt{q} \times \frac{6 q^{2} (p - o) + (14 o p - 5 (o^{2} + p^{2})) q - 6 o p (p - o)}{(2 q - (p - o)) \sqrt{2 (p - o)}}$ $f^{'} (q) = 0 \land q \neq 0 \Rightarrow$ $\Rightarrow q = \frac{5 o^{2} - 14 o p + 5 p^{2}}{12 (p - o)} \pm \frac{(o + p) \sqrt{25 o^{2} - 46 o p + 25 p^{2}}}{12 (p - o)}$ $trying with o = 1 and p > 1 \Rightarrow we need the$ $solution with the +$ $so we have$ $a^{2} = \frac{2 q^{2} (2 o p + q (p - o))}{(p - o) (2 q - (p - o))}$ $b^{2} = \frac{q (2 o p + q (p - o))}{2 q - (p - o)}$ $q = \frac{5 o^{2} - 14 o p + 5 p^{2}}{12 (p - o)} + \frac{(o + p) \sqrt{25 o^{2} - 46 o p + 25 p^{2}}}{12 (p - o)}$
	Commented by MJS last updated on 20/Nov/18

	$but {there}^{'} s a minimum value for q depending$ $on o and p so that the ellipse is touching the$ $1^{st} circle : q ⩾ \frac{2 o (p - o)}{3 o - p} . above formula for q$ $meets this at o < p ⩽ \frac{3 o}{2} . p > \frac{3 o}{2} \Rightarrow smallest$ $ellipse with q = \frac{2 o (p - o)}{3 o - p} which indeed makes$ $the smaller circle the osculating circle of the$ $ellipse when p > \frac{3 o}{2}$
	Commented by ajfour last updated on 20/Nov/18

	$T h a n k y o u f o r a s m u c h, S i r!$
	Answered by ajfour last updated on 20/Nov/18

	$L e t p o i n t o f c o n t a c t o f t h e t w o$ $c i r c l e s b e t h e o r i g i n O .$ $\underline{E q . o f e l l i p s e i s}$ $\frac{{(x - h)}^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $\underline{F o r g e n e r a l e q . f o r b o t h c i r c l e s}$ ${(x - ρ)}^{2} + y^{2} = ρ^{2}$ $F o r t a n g e n c y c o n d i t i o n, s u c h a$ $c i r c l e a n d e l l i p s e h a v e o n e c o m m o n$ $r o o t . \Rightarrow e q . b e l o w h a s d o u b l e r o o t .$ $\frac{{(x - h)}^{2}}{a^{2}} + \frac{ρ^{2} - {(x - ρ)}^{2}}{b^{2}} = 1$ $o r$ $x^{2} (\frac{1}{a^{2}} - \frac{1}{b^{2}}) - 2 x (\frac{h}{a^{2}} - \frac{ρ}{b^{2}}) + \frac{h^{2}}{a^{2}} - 1 = 0$ $\Rightarrow 4 {(\frac{h}{a^{2}} - \frac{ρ}{b^{2}})}^{2} = 4 (\frac{1}{b^{2}} - \frac{1}{a^{2}}) (1 - \frac{h^{2}}{a^{2}})$ $\Rightarrow \frac{ρ^{2}}{b^{4}} - \frac{2 ρ h}{a^{2} b^{2}} = (\frac{1}{b^{2}} - \frac{1}{a^{2}}) - \frac{h^{2}}{a^{2} b^{2}}$ $A b o v e e q . i n ρ h a s t w o r o o t s$ $ρ = R a n d ρ = - r$ $\Rightarrow {(R - r)}^{2} = \frac{4 h^{2} b^{4}}{a^{4}} &$ $R r = b^{4} [(\frac{1}{b^{2}} - \frac{1}{a^{2}}) - \frac{h^{2}}{a^{2} b^{2}}]$ $\Rightarrow h^{2} = \frac{a^{4} {(R - r)}^{2}}{4 b^{4}} a n d a l s o$ $h^{2} = a^{2} - b^{2} - \frac{a^{2} R r}{b^{2}}$ $\Rightarrow \frac{a^{4} {(R - r)}^{2}}{4 b^{4}} = a^{2} - b^{2} - \frac{a^{2} R r}{b^{2}}$ $\Rightarrow \frac{a^{2} {(R - r)}^{2}}{4} = b^{4} - \frac{b^{6}}{a^{2}} - b^{2} R r$ $A n d a r e a o f e l l i p s e = π a b$ $l e t a^{2} b^{2} = p \Rightarrow a^{2} = \frac{p}{b^{2}}$ $f u r t h e r l e t b^{2} = t \Rightarrow a^{2} = \frac{p}{t}$ $\frac{p {(R - r)}^{2}}{4 t} = t^{2} - \frac{t^{4}}{p} - R r t . . . (I)$ $F o r m i n i m u m a r e a p i s m i n i m u m$ $T h e r e f o r e \frac{d p}{d t} = 0$ $\Rightarrow - \frac{p {(R - r)}^{2}}{4 t^{2}} = 2 t - \frac{4 t^{3}}{p} - R r . . (I I)$ $(I) + t \times (I I) y i e l d s$ $0 = 3 t^{2} - \frac{5 t^{4}}{p} - 2 R r t$ $S o p = \frac{5 t^{3}}{3 t - 2 R r} . . . . (i)$ $N o w \frac{4}{t} \times (I) - (I I) g i v e s$ $\frac{5 p {(R - r)}^{2}}{4 t^{2}} = 2 t - 3 R r$ $u s i n g (i) i n a b o v e e q .$ $\frac{25 t {(R - r)}^{2}}{4 (3 t - 2 R r)} = 2 t - 3 R r$ $\Rightarrow l e t z = \frac{t}{R r} = \frac{b^{2}}{R r}, t h e n$ $\frac{25 z {(R - r)}^{2}}{4 R r (3 z - 2)} = 2 z - 3$ $A l s o l e t \frac{5 (R - r)}{2 \sqrt{R r}} = c, t h e n$ $(3 z - 2) (2 z - 3) = c^{2} z$ $\Rightarrow 6 z^{2} - (c^{2} + 13) z + 6 = 0$ $z = \frac{c^{2} + 13 \pm \sqrt{{(c^{2} + 13)}^{2} - 144}}{12}$ $\Rightarrow \frac{b^{2}}{R r} = \frac{25}{48} \frac{{(R - r)}^{2}}{R r} + \frac{13}{12}$ $\pm \sqrt{{(\frac{25}{48} \frac{{(R - r)}^{2}}{R r} + \frac{13}{12})}^{2} - 1}$ $I f r = R$ $\frac{b}{R} = \sqrt{\frac{13}{12} \pm \frac{5}{12}}$ $b u t \frac{b}{R} > 1 \Rightarrow + s i g n g i v e s$ $c o r r e c t r e s u l t . b = \sqrt{\frac{3}{2}} R$ $A n d g e n e r a l l y$ $\frac{b^{2}}{R r} = \frac{25 {(R - r)}^{2}}{48 R r} + \frac{13}{12}$ $+ \sqrt{{(\frac{25 {(R - r)}^{2}}{48 R r} + \frac{13}{12})}^{2} - 1}$ $u s i n g t h i s v a l u e i n$ $p = \frac{5 t^{3}}{3 t - 2 R r} g i v e s$ $a^{2} b^{2} = \frac{5 b^{6}}{3 b^{2} - 2 R r}$ $\Rightarrow \frac{a^{2}}{R r} = \frac{5 {(\frac{b^{2}}{R r})}^{2}}{3 (\frac{b^{2}}{R r}) - 2}$ $A s a s p e c i a l c a s e f o r r = R$ $b = \frac{\sqrt{3}}{\sqrt{2}} R;$ $\Rightarrow \frac{a^{2}}{R^{2}} = \frac{5 (\frac{9}{4})}{3 (\frac{3}{2}) - 2} = \frac{9}{2}$ $\Rightarrow a = \frac{3}{\sqrt{2}} R, b = \frac{\sqrt{3}}{\sqrt{2}} R$ $_________________________.$
	Commented by MJS last updated on 20/Nov/18

	$my solution includes the same issue .$ $before trying to find the minimum of a b we$ $have q as parameter . obviously there are$ $borders for q depending on the values of$ $o and p (r and R)$
	Commented by mr W last updated on 20/Nov/18

	$v e r y n i c e m e t h o d s i r!$ $b u t i t s e e m s t o w o r k o n l y f o r c a s e s$ $w h e r e R a n d r a r e n o t v e r y d i f f e r e n t .$ $F o r r < R / 2 t h e r e s u l t s s e e m n o t t o b e$ $c o r r e c t .$ $I a m s h o w i n g t w o c a s e s :$ $r = 6, R = 8 a n d r = 3, R = 8$
	Commented by mr W last updated on 20/Nov/18

	Commented by mr W last updated on 20/Nov/18

	Commented by ajfour last updated on 20/Nov/18

	$H o w c o m e t h e a m b i g u i t y, S i r ?$ $p l e a s e h e l p . . . .$
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