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Question Number 50048 by ajfour last updated on 13/Dec/18

Commented by ajfour last updated on 13/Dec/18

$F i n d R i n t e r m s o f a a n d b .$
Commented by ajfour last updated on 13/Dec/18

Commented by MJS last updated on 14/Dec/18

$the problem can be seen as follows :$ $2 circles are given with radii a and b, they$ $have exactly 1 common point$ ${we}^{'} re looking (1) for a 3^{rd} circle touching both$ $in exactly 1 common point each, with radius R$ ${we}^{'} re looking (2) for pairs of tangents to$ $circles a and b which include a right angle$ $the goal is to find R and the tangents meeting$ $in the center of the 3^{rd} circle$ $we can put the circles as follows :$ ${(x - a)}^{2} + y^{2} = a^{2}; a < 0$ ${(x - b)}^{2} + y^{2} = b^{2}; b > 0$ ${(x - p)}^{2} + {(y - q)}^{2} = R^{2}$ $\Rightarrow p = \frac{a + b}{a - b} R; q = - \frac{2}{a - b} \sqrt{a b R (a - b - R)}$ $we need the upper half circles for the$ $tangents . let x = u for circle a and x = v for$ $circle b \Rightarrow tangent at x = v normal to tangent$ $in x = u \Leftrightarrow v = b - \frac{b}{a} \sqrt{2 a u - u^{2}}$ $the tangents then intersect at (\begin{matrix} m \\ n \end{matrix}) with$ $m = \frac{a - b}{a^{2}} u^{2} + \frac{2 b - a}{a} u - \frac{b}{a} \sqrt{2 a u - u^{2}}$ $n = \frac{b}{a} u - b + \frac{(a - b) u + a b}{a^{2}} \sqrt{2 a u - u^{2}}$ $now we have the system$ $p = m$ $q = n$ $which must be solved for u and R$ $but it leads to a polynome of 8^{th} degree . . .$
Commented by ajfour last updated on 14/Dec/18

$T h a n k s f o r t h e o v e r v i e w S i r, i$ $m y s e l f w a n t e d t o a n a l y s e i t t h i s$ $w a y .$
	Answered by behi83417@gmail.com last updated on 13/Dec/18

	$c o s α = \frac{\sqrt{{(R + b)}^{2} - b^{2}}}{R + b} = \frac{\sqrt{R^{2} + 2 R b}}{R + b} = \frac{\sqrt{1 + 2 n}}{1 + n}$ $c o s γ = \frac{\sqrt{{(R + a)}^{2} - a^{2}}}{R + a} = \frac{\sqrt{R^{2} + 2 R a}}{R + a} = \frac{\sqrt{1 + 2 m}}{1 + m}$ $s i n α = \frac{b}{R + b} = \frac{n}{1 + n}, s i n γ = \frac{a}{R + a} = \frac{m}{1 + m}$ $\frac{a}{R} = m, \frac{b}{R} = n$ $s i n (α + γ) = \frac{m \sqrt{1 + 2 n} + n \sqrt{1 + 2 m}}{(1 + m) (1 + n)}$ $= \frac{s (t^{2} - 1) + t (s^{2} - 1)}{4 (1 + \frac{t^{2} - 1}{2}) (1 + \frac{s^{2} - 1}{2})} = \frac{(s t - 1) (s + t)}{(s^{2} + 1) (t^{2} + 1)}$ $[1 + 2 m = t^{2}, 1 + 2 n = s^{2} \Rightarrow m = \frac{t^{2} - 1}{2}, n = \frac{s^{2} - 1}{2}]$ $β = 90 - (α + γ) \Rightarrow c o s β = s i n (α + γ)$ ${(a + b)}^{2} = {(R + a)}^{2} + {(R + b)}^{2} - 2 (R + a) (R + b) . c o s β \Rightarrow$ ${(m + n)}^{2} = [{(1 + m)}^{2} + {(1 + n)}^{2} - 2 (m \sqrt{1 + 2 n} + n \sqrt{1 + 2 m})]$ ${(\frac{t^{2} + s^{2} - 2}{2})}^{2} = [{(\frac{t^{2} + 1}{2})}^{2} + {(\frac{s^{2} + 1}{2})}^{2} - 2 (\frac{t^{2} - 1}{2} s + \frac{s^{2} - 1}{2} t)]$ ${(t^{2} + s^{2} - 2)}^{2} = {(t^{2} + 1)}^{2} + {(s^{2} + 1)}^{2} - 4 ({s t}^{2} + {t s}^{2} - s - t)$ $t^{4} + s^{4} + 4 - 4 t^{2} - 4 s^{2} + 2 t^{2} s^{2} =$ $t^{4} + 2 t^{2} + 1 + s^{4} + 2 s^{2} + 1 - 4 {s t}^{2} - 4 {t s}^{2} + 4 s + 4 t$ $3 t^{2} + 3 s^{2} - t^{2} s^{2} - 2 {s t}^{2} - 2 {t s}^{2} + 2 s + 2 t = 0$ $(s^{2} + 2 s - 3) t^{2} + 2 (s^{2} - 1) t - (3 s^{2} + 2 s) = 0$ $△^{'} = b^{' 2} - a c = {(s^{2} - 1)}^{2} + (s^{2} + 2 s - 3) (3 s^{2} + 2 s) =$ $= s^{4} - 2 s^{2} + 1 + 3 s^{4} + 2 s^{3} + 6 s^{3} + 4 s^{2} - 9 s^{2} - 6 s =$ $= 4 s^{4} + 8 s^{3} - 7 s^{2} - 6 s + 1$ $= (s - 1) [4 s^{3} + 12 s^{2} + 5 s - 1]$ $t = \frac{(1 - s^{2}) \pm \sqrt{4 s^{4} + 8 s^{3} - 7 s^{2} - 6 s + 1}}{s^{2} + 2 s - 3}$ $. . . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 13/Dec/18
	$eq of circles x^2 +y^2 =R^2 (x−α)^2 +(y−b^2 )=b^2 (x−a)^2 +(y−β)^2 =a^2 (R+b)^2 =(α−0)^2 +(b−0)^2 R^2 +2Rb−α^2 =0 (R+a)^2 =(a−0)^2 +(β−0)^2 R^2 +2Ra−β^2 =0 m_1 =tanθ_1 =((b−0)/(α−0))=(b/(√(R^2 +2Rb))) m_2 =tanθ_2 =((β−0)/(a−0))=((√(R^2 +2Ra))/a) cos(θ_2 −θ_1 )=((R^2 +R^2 −(a+b)^2 )/(2.R.R)) 2R^2 {1−cos(θ_2 −θ_1 )}=(a+b)^2 2Rsin(((θ_2 −θ_1 )/2))=(a+b) sin(((θ_2 −θ_ )/2))=((a+b)/(2R)) tan{2×(((θ_2 −θ_1 )/2))}=tan(θ_2 −θ_1 ) ((2tan(((θ_2 −θ_1 )/2)))/(1−tan^2 (((θ_2 −θ_1 )/2))))=((tanθ_2 −tanθ_1 )/(1+tanθ_2 tanθ_ )) ((2×(((a+b))/(√(4R^2 −(a+b)^2 ))))/(1−(((a+b)^2 )/(4R^2 −(a+b)^2 )))) =((((√(R^2 +2Ra))/a)−(b/(√(R^2 +2Rb))))/(1+(b/a)×(√((R^2 +2Ra)/(R^2 +2Rb))))) wait i have to simplify...$
	$e q o f c i r c l e s$ $x^{2} + y^{2} = R^{2}$ ${(x - α)}^{2} + (y - b^{2}) = b^{2}$ ${(x - a)}^{2} + {(y - β)}^{2} = a^{2}$ ${(R + b)}^{2} = {(α - 0)}^{2} + {(b - 0)}^{2}$ $R^{2} + 2 R b - α^{2} = 0$ ${(R + a)}^{2} = {(a - 0)}^{2} + {(β - 0)}^{2}$ $R^{2} + 2 R a - β^{2} = 0$ $m_{1} = t a n θ_{1} = \frac{b - 0}{α - 0} = \frac{b}{\sqrt{R^{2} + 2 R b}}$ $m_{2} = t a n θ_{2} = \frac{β - 0}{a - 0} = \frac{\sqrt{R^{2} + 2 R a}}{a}$ $c o s (θ_{2} - θ_{1}) = \frac{R^{2} + R^{2} - {(a + b)}^{2}}{2 . R . R}$ $2 R^{2} {1 - c o s (θ_{2} - θ_{1})} = {(a + b)}^{2}$ $2 R s i n (\frac{θ_{2} - θ_{1}}{2}) = (a + b)$ $s i n (\frac{θ_{2} - θ_{}}{2}) = \frac{a + b}{2 R}$ $t a n {2 \times (\frac{θ_{2} - θ_{1}}{2})} = t a n (θ_{2} - θ_{1})$ $\frac{2 t a n (\frac{θ_{2} - θ_{1}}{2})}{1 - {t a n}^{2} (\frac{θ_{2} - θ_{1}}{2})} = \frac{t a n θ_{2} - t a n θ_{1}}{1 + t a n θ_{2} t a n θ_{}}$ $\frac{2 \times \frac{(a + b)}{\sqrt{4 R^{2} - {(a + b)}^{2}}}}{1 - \frac{{(a + b)}^{2}}{4 R^{2} - {(a + b)}^{2}}} = \frac{\frac{\sqrt{R^{2} + 2 R a}}{a} - \frac{b}{\sqrt{R^{2} + 2 R b}}}{1 + \frac{b}{a} \times \sqrt{\frac{R^{2} + 2 R a}{R^{2} + 2 R b}}}$ $w a i t i h a v e t o s i m p l i f y . . .$
	Answered by ajfour last updated on 13/Dec/18

	$l e t \frac{b}{R} = s, \frac{a}{R} = t$ $\sin α = \frac{b}{R + b} = \frac{s}{1 + s}$ $\sin γ = \frac{a}{R + a} = \frac{t}{1 + t}$ $\cos β = \frac{{(R + a)}^{2} + {(R + b)}^{2} - {(a + b)}^{2}}{2 (R + a) (R + b)}$ $= \frac{{(1 + t)}^{2} + {(1 + s)}^{2} - {(t + s)}^{2}}{2 (1 + t) (1 + s)}$ $= \frac{1 + t + s - s t}{1 + t + s + s t}$ $\cos β = \sin (α + γ)$ $\Rightarrow \frac{1 + t + s - s t}{1 + t + s + s t} = \frac{s}{1 + s} \times \frac{\sqrt{1 + 2 t}}{1 + t} + \frac{t}{1 + t} \times \frac{\sqrt{1 + 2 s}}{1 + s}$ $\Rightarrow 1 + s + t - s t = s \sqrt{1 + 2 t} + t \sqrt{1 + 2 s}$ $\Rightarrow 1 + \frac{b}{R} + \frac{a}{R} - \frac{a b}{R^{2}} = \frac{b}{R} \sqrt{1 + \frac{2 a}{R}} + \frac{a}{R} \sqrt{1 + \frac{2 b}{R}}$ $R^{2} + (a + b) R - a b = \sqrt{R} (b \sqrt{R + 2 a} + a \sqrt{R + 2 b})$ $l e t \frac{R}{b} = x, \frac{a}{b} = c \Rightarrow$ $x^{2} + (c + 1) x - c = \sqrt{x} (\sqrt{x + 2 c} + c \sqrt{x + 2})$ $\Rightarrow i c a n n o t s o l v e i t .$
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