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Question Number 53530 by MJS last updated on 23/Jan/19

Commented by MJS last updated on 23/Jan/19

$I found a solution to question 53168 by$ $Sir Aifour$
	Answered by MJS last updated on 23/Jan/19

	$I changed the setup after some failed attempts$ $I put a = 1; b = a^{2}; \frac{1}{3} < b ⩽ 1 \Leftrightarrow \frac{\sqrt{3}}{3} < a ⩽ 1$ $to go on with b leads to more complications$ $because of terms with \sqrt{b^{2 n - 1}}; the borders$ $for a make sure both circles are within the$ $triangle (we could allow \frac{1}{3} < b < 3 leading to$ $symmetric results)$ $M = (\begin{matrix} - 1 \\ 0 \end{matrix}) N = (\begin{matrix} a^{2} \\ 0 \end{matrix})$ $the given circles$ ${(x + 1)}^{2} + y^{2} = 1$ ${(x - a^{2})}^{2} + y^{2} = a^{4}$ $now {it}^{'} s possible to get the equations of the$ $lines B C, C A & A B :$ $B C : y = \frac{1 - a^{2}}{2 a} x - a$ $C A : \frac{a^{2} + 2 \sqrt{3} a - 1}{\sqrt{3} a^{2} - 2 a - \sqrt{3}} x - \frac{a^{2} (\sqrt{3} a + 1)}{a - \sqrt{3}}$ $A B : - \frac{\sqrt{3} a^{2} - 6 a - \sqrt{3}}{3 a^{2} + 2 \sqrt{3} a - 3} x + \frac{\sqrt{3} a + 3}{3 a - \sqrt{3}}$ $A = (\begin{matrix} \frac{(a - 1) (a + 1) (3 a^{2} + 4 \sqrt{3} a + 3)}{2 (a^{2} + 1)} \\ - \frac{3 \sqrt{3} a^{5} - 15 a^{4} - 14 \sqrt{3} a^{3} + 6 a^{2} + 7 \sqrt{3} a - 3}{2 (3 a - \sqrt{3}) (a^{2} + 1)} \end{matrix})$ $B = (\begin{matrix} - \frac{2 a (a + \sqrt{3})}{a^{2} + 1} \\ \frac{\sqrt{3} (\sqrt{3} a - 3) (\sqrt{3} a + 1)}{3 (a^{2} + 1)} \end{matrix})$ $C = (\begin{matrix} \frac{2 a^{2} (\sqrt{3} a + 1)}{a^{2} + 1} \\ - \frac{a^{2} (\sqrt{3} a^{2} + 2 a - \sqrt{3})}{a^{2} + 1} \end{matrix})$ $[the roots of 3 \sqrt{3} a^{5} - 15 a^{4} - 14 \sqrt{3} a^{3} + 6 a^{2} + 7 \sqrt{3} a - 3 :$ $a_{1} = \frac{\sqrt{3}}{3}; a_{2, 3} = \frac{\sqrt{3}}{3} (1 + \sqrt{7} \pm \sqrt{5 + 2 \sqrt{7}}); a_{4, 5} = \frac{\sqrt{3}}{3} (1 - \sqrt{7} \pm \sqrt{- 5 + 2 \sqrt{7}} i)]$ $the side of the triangle$ $s = \sqrt{3} a^{2} + 2 a + \sqrt{3}$ $the searched circle can be found like this$ $P = (\begin{matrix} p \\ q \end{matrix})$ ${(x - p)}^{2} + {(y - q)}^{2} = r^{2}$ $steps 1 & 2 : intersect with the given circles$ $so that {there}^{'} s exactly one common point$ $with each of them \Rightarrow p, q$ $step 3 : A \in circle \Rightarrow r$ $these steps lead to 2 solutions, one of the$ $found circles i n c l u d e s the given circles .$ $the hardest work is to simplify the polynomes$ $I spent a day searching for zeros . . .$ $p = - \frac{a^{2} - 1}{a^{2} + 1} r$ $q = \frac{2 a \sqrt{r^{2} + (a^{2} + 1) r}}{a^{2} + 1}$ $r = - \frac{9 {(a^{2} + (\frac{\sqrt{3}}{6} + \frac{\sqrt{11}}{2}) a + 1)}^{2} {(a^{2} + (\frac{\sqrt{3}}{6} - \frac{\sqrt{11}}{2}) a + 1)}^{2}}{(a - \sqrt{3}) (a + \sqrt{3}) (\sqrt{3} a - 1) (\sqrt{3} a + 1) (3 a^{2} + 2 \sqrt{3} a + 3)}$ ${isn}^{'} t it of a strange but great beauty ?$
	Commented by ajfour last updated on 23/Jan/19

	$i t h i n k, i t s j u s t a q u a d r a t i c i n R; o n l y$ $q u i t e c o m p l i c a t e d . t h a n k s S i r .$
	Commented by MJS last updated on 23/Jan/19
	$it is indeed quadratic in r (your R) but it′s very complicated to solve because it′s degree 12 in a and we must constantly find zeros to shorten huge fractions. in the end everything is fine. even the root in my q vanishes because it′s (√(term(a)^2 ))... I′ll post later$
	$it is indeed quadratic in r (your R) but {it}^{'} s$ $very complicated to solve because {it}^{'} s degree$ $12 in a and we must constantly find zeros$ $to shorten huge fractions . in the end$ $everything is fine . even the root in my$ $q vanishes because {it}^{'} s \sqrt{t e r m {(a)}^{2}} . . . I^{'} ll$ $post later$
	Answered by ajfour last updated on 23/Jan/19

	$l e t M b e o r i g i n . x - a x i s ∥ B C$ $s = \sqrt{3} (a + b) + 2 \sqrt{a b}$ $L e t P (h, k), N (2 \sqrt{a b}, b - a)$ $A (\frac{s}{2} - a \sqrt{3}, \frac{s \sqrt{3}}{2} - a) .$ $h^{2} + k^{2} = {(a + R)}^{2} . . . (i)$ ${(h - 2 \sqrt{a b})}^{2} + {(k - b + a)}^{2} = {(b + R)}^{2} . . (i i)$ ${(h - \frac{s}{2} + a \sqrt{3})}^{2} + {(k - \frac{s \sqrt{3}}{2} + a)}^{2} = R^{2} . . (i i i)$ $(i i) - (i), (i i i) - (i) g i v e s$ $A h + B k = (b - a) (a + b + R)$ $C h + D k = - a (2 a + R)$ $A, B, C, D c o n t a i n s o n l y a a n d b .$ $U p o n f i n d i n g (h, k) w e u s e e q . (i)$ $I f i l e t \frac{b}{a} = t^{2}, \frac{h}{a} = p, \frac{k}{a} = q, \frac{R}{a} = ρ$ $t h e n \frac{s}{2 a} = \sqrt{3} (1 + t^{2}) + 2 t = u$ $p^{2} + q^{2} = {(1 + ρ)}^{2} . . . (I)$ ${(p - 2 t)}^{2} + {(q - t^{2} + 1)}^{2} = {(t^{2} + ρ)}^{2} . . . (I I)$ ${(p - u + \sqrt{3})}^{2} + {(q - u \sqrt{3} + 1)}^{2} = ρ^{2} . . (I I I)$ $(I I) - (I) & (I I I) - (I) g i v e$ $- 4 t p + 4 t^{2} - 2 (t^{2} - 1) q + {(t^{2} - 1)}^{2} = (t^{2} + 1) (1 + t^{2} + 2 ρ)$ $&$ $- 2 (u - \sqrt{3}) p + {(u - \sqrt{3})}^{2} - 2 (u \sqrt{3} - 1) q$ $+ {(u \sqrt{3} - 1)}^{2} = - (2 ρ + 1)$
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