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Question Number 38092 by ajfour last updated on 21/Jun/18

Commented by ajfour last updated on 22/Jun/18

$T h e c i r c l e t o u c h e s x = 0, y = 0,$ $a n d t h e e l l i p s e \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, i n t h e$ $m a n n e r s h o w n; f i n d i t s r a d i u s R$ $i n t e r m s o f a a n d b .$
Commented by MJS last updated on 22/Jun/18

$there are 2 circles : the one you show and a$ $small one within the ellipse .$ $I^{'} m afraid this leads to polynomes of 4^{th} or$ $even 6^{th} degree . . .$
Commented by MJS last updated on 22/Jun/18
$after not succeeding from the front I tried it from the rear: start with the common tangent in the intersection point P t: y=kx+d; k<0; d>0 P∈t: P= ((p),((kp+d)) ) n⊥t; P∈n n: y=−(1/k)x+d+(((k^2 +1)p)/k) the centers of the circles lie on m: y=x m∩n x=−(1/k)x+d+(((k^2 +1)p)/k) ⇒ x=((dk+(k^2 +1)p)/(k+1)) C= ((((dk+(k^2 +1)p)/(k+1))),(((dk+(k^2 +1)p)/(k+1))) ) ∣CP∣=r=x (((dk+(k^2 +1)p)/(k+1))−p)^2 +(((dk+(k^2 +1)p)/(k+1))−kp−d)^2 −(((dk+(k^2 +1)p)/(k+1)))^2 =0 ⇒ { ((p_1 =(d/(2k))(−1−((k+1)/(√(k^2 +1)))))),((p_2 =(d/(2k))(−1+((k+1)/(√(k^2 +1)))))) :} ⇒ { ((P_1 = ((((d/(2k))(−1−((k+1)/(√(k^2 +1)))))),(((d/2)(1−((k+1)/(√(k^2 +1)))))) ))),((P_2 = ((((d/(2k))(−1+((k+1)/(√(k^2 +1)))))),(((d/2)(1+((k+1)/(√(k^2 +1)))))) ))) :} ⇒ { ((C_1 = ((r_1 ),(r_1 ) ); r_1 =(d/(2k))(k−1−(√(k^2 +1))))),((C_2 = ((r_2 ),(r_2 ) ); r_2 =(d/(2k))(k−1+(√(k^2 +1))))) :} now we can find 2 ellipses for given k and d t_P ^(ell) =−((bp)/(a(√(a^2 −p^2 ))))x+((ab)/(√(a^2 −p^2 ))) = kx+d ⇒ k=−((bp)/(a(√(a^2 −p^2 )))); d=((ab)/(√(a^2 −p^2 ))) ⇒ a=(√(dp/(−k))); b=(√(d(d+kp)))$
$after not succeeding from the front I tried$ $it from the rear :$ $start with the common tangent in the$ $intersection point P$ $t : y = k x + d; k < 0; d > 0$ $P \in t : P = (\begin{matrix} p \\ k p + d \end{matrix})$ $n ⊥ t; P \in n$ $n : y = - \frac{1}{k} x + d + \frac{(k^{2} + 1) p}{k}$ $the centers of the circles lie on$ $m : y = x$ $m \cap n$ $x = - \frac{1}{k} x + d + \frac{(k^{2} + 1) p}{k} \Rightarrow x = \frac{d k + (k^{2} + 1) p}{k + 1}$ $C = (\begin{matrix} \frac{d k + (k^{2} + 1) p}{k + 1} \\ \frac{d k + (k^{2} + 1) p}{k + 1} \end{matrix})$ $∣ C P ∣= r = x$ ${(\frac{d k + (k^{2} + 1) p}{k + 1} - p)}^{2} + {(\frac{d k + (k^{2} + 1) p}{k + 1} - k p - d)}^{2} - {(\frac{d k + (k^{2} + 1) p}{k + 1})}^{2} = 0$ $\Rightarrow {\begin{cases} p_{1} = \frac{d}{2 k} (- 1 - \frac{k + 1}{\sqrt{k^{2} + 1}}) \\ p_{2} = \frac{d}{2 k} (- 1 + \frac{k + 1}{\sqrt{k^{2} + 1}}) \end{cases}$ $\Rightarrow {\begin{cases} P_{1} = (\begin{array}{c} \frac{d}{2 k} (- 1 - \frac{k + 1}{\sqrt{k^{2} + 1}}) \\ \frac{d}{2} (1 - \frac{k + 1}{\sqrt{k^{2} + 1}}) \end{array}) \\ P_{2} = (\begin{array}{c} \frac{d}{2 k} (- 1 + \frac{k + 1}{\sqrt{k^{2} + 1}}) \\ \frac{d}{2} (1 + \frac{k + 1}{\sqrt{k^{2} + 1}}) \end{array}) \end{cases}$ $\Rightarrow {\begin{cases} C_{1} = (\begin{array}{c} r_{1} \\ r_{1} \end{array}); r_{1} = \frac{d}{2 k} (k - 1 - \sqrt{k^{2} + 1}) \\ C_{2} = (\begin{array}{c} r_{2} \\ r_{2} \end{array}); r_{2} = \frac{d}{2 k} (k - 1 + \sqrt{k^{2} + 1}) \end{cases}$ $now we can find 2 ellipses for given k and d$ $t_{P}^{ell} = - \frac{b p}{a \sqrt{a^{2} - p^{2}}} x + \frac{a b}{\sqrt{a^{2} - p^{2}}} = k x + d$ $\Rightarrow k = - \frac{b p}{a \sqrt{a^{2} - p^{2}}}; d = \frac{a b}{\sqrt{a^{2} - p^{2}}}$ $\Rightarrow a = \sqrt{\frac{d p}{- k}}; b = \sqrt{d (d + k p)}$
Commented by ajfour last updated on 23/Jun/18

$E x c e l l e n t! S i r .$ $a i n t y o u r e x p r e s s i o n s f o r p_{1}, p_{2}$ $a b i t s i m p l e r; i d o n t g e t t h e s a m e . .$
Commented by MJS last updated on 25/Jun/18

${(\frac{d k + (k^{2} + 1) p}{k + 1} - p)}^{2} + {(\frac{d k + (k^{2} + 1) p}{k + 1} - k p - d)}^{2} - {(\frac{d k + (k^{2} + 1) p}{k + 1})}^{2} = 0$ $\frac{{(d k + k (k - 1) p)}^{2} + {(- d - (k - 1) p)}^{2} - {(d k + (k^{2} + 1) p)}^{2}}{{(k + 1)}^{2}} = 0$ $k^{2} {(k - 1)}^{2} p^{2} + 2 {d k}^{2} (k - 1) p + d^{2} k^{2} +$ $+ {(k - 1)}^{2} p^{2} + 2 d (k - 1) p + d^{2} -$ $- ({(k^{2} + 1)}^{2} p^{2} + 2 d k (k^{2} + 1) p + d^{2} k^{2}) = 0$ $- 2 k (k^{2} + 1) p^{2} - 2 d (k^{2} + 1) p + d^{2} = 0$ $p^{2} + \frac{d}{k} p - \frac{d^{2}}{2 k (k^{2} + 1)} = 0$ $p = - \frac{d}{2 k} \pm \sqrt{\frac{d^{2}}{4 k^{2}} + \frac{d^{2}}{2 k (k^{2} + 1)}} =$ $= - \frac{d}{2 k} \pm \sqrt{\frac{d^{2} (k^{2} + 1) + 2 d^{2} k}{4 k^{2} (k^{2} + 1)}} =$ $= - \frac{d}{2 k} \pm \frac{d}{2 k} \sqrt{\frac{k^{2} + 1 + 2 k}{k^{2} + 1}} =$ $= \frac{d}{2 k} (- 1 \pm \frac{k + 1}{\sqrt{k^{2} + 1}})$
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