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Question Number 183295 by ajfour last updated on 24/Dec/22

Commented by ajfour last updated on 24/Dec/22

$F i n d R i n t e r m s o f a a n d b .$
	Answered by ajfour last updated on 24/Dec/22
	$let common tangent x axis. Where big circle touches it be origin. A(−p,0) ; B(q,0) (p+q)^2 =4ab k=b+(((a−b)/(a+b)))b=((2ab)/(a+b)) h=−p+((a(p+q))/(a+b))=((aq−bp)/(a+b)) M(h,−k)≡(((aq−bp)/(a+b)), −((2ab)/(a+b))) Eq. of big circle: x^2 +(y−R)^2 =R^2 eq. of left tangent to big circle y=−(x+p){(((((2ab)/(a+b))))/((a(p+q))/(a+b)))} ⇒ y=−(x+p)(((2b)/(p+q))) eq. of right tangent to big circle y=(x−q){(((((2ab)/(a+b))))/((b(p+q))/(a+b)))}=(x−q)(((2a)/(p+q))) R^2 =(((R+((2bp)/(p+q)))^2 )/(1+((4b^2 )/((p+q)^2 ))))=(((R+((2aq)/(p+q)))^2 )/(1+((4a^2 )/((p+q)^2 )))) but (p+q)^2 =4ab ⇒ R^2 =(((2R(√(ab))+2bp)^2 )/(4b(a+b)))=(((2R(√(ab))+2aq)^2 )/(4a(a+b))) ⇒ (R/( (√b)))((√(a+b))−(√a))=p (R/( (√a)))((√(a+b))−(√b))=q adding R{(√((a/b)+1))−(√(a/b))+(√(1+(b/a)))−(√(b/a))} =2(√(ab)) R=((2ab)/( ((√a)+(√b))(√(a+b))−(a+b)))$
	$l e t c o m m o n t a n g e n t x a x i s .$ $W h e r e b i g c i r c l e t o u c h e s i t b e$ $o r i g i n .$ $A (- p, 0); B (q, 0)$ ${(p + q)}^{2} = 4 a b$ $k = b + (\frac{a - b}{a + b}) b = \frac{2 a b}{a + b}$ $h = - p + \frac{a (p + q)}{a + b} = \frac{a q - b p}{a + b}$ $M (h, - k) \equiv (\frac{a q - b p}{a + b}, - \frac{2 a b}{a + b})$ $E q . o f b i g c i r c l e : x^{2} + {(y - R)}^{2} = R^{2}$ $e q . o f l e f t t a n g e n t t o b i g c i r c l e$ $y = - (x + p) {\frac{(\frac{2 a b}{a + b})}{\frac{a (p + q)}{a + b}}}$ $\Rightarrow y = - (x + p) (\frac{2 b}{p + q})$ $e q . o f r i g h t t a n g e n t t o b i g c i r c l e$ $y = (x - q) {\frac{(\frac{2 a b}{a + b})}{\frac{b (p + q)}{a + b}}} = (x - q) (\frac{2 a}{p + q})$ $R^{2} = \frac{{(R + \frac{2 b p}{p + q})}^{2}}{1 + \frac{4 b^{2}}{{(p + q)}^{2}}} = \frac{{(R + \frac{2 a q}{p + q})}^{2}}{1 + \frac{4 a^{2}}{{(p + q)}^{2}}}$ $b u t {(p + q)}^{2} = 4 a b \Rightarrow$ $R^{2} = \frac{{(2 R \sqrt{a b} + 2 b p)}^{2}}{4 b (a + b)} = \frac{{(2 R \sqrt{a b} + 2 a q)}^{2}}{4 a (a + b)}$ $\Rightarrow \frac{R}{\sqrt{b}} (\sqrt{a + b} - \sqrt{a}) = p$ $\frac{R}{\sqrt{a}} (\sqrt{a + b} - \sqrt{b}) = q$ $a d d i n g$ $R {\sqrt{\frac{a}{b} + 1} - \sqrt{\frac{a}{b}} + \sqrt{1 + \frac{b}{a}} - \sqrt{\frac{b}{a}}}$ $= 2 \sqrt{a b}$ $R = \frac{2 a b}{(\sqrt{a} + \sqrt{b}) \sqrt{a + b} - (a + b)}$
	Commented by mr W last updated on 24/Dec/22

	$i t c a n b e s i m p l i f i e d t o$ $R = \frac{\sqrt{a b} (\sqrt{a} + \sqrt{b} + \sqrt{a + b})}{\sqrt{a + b}}$
	Answered by mr W last updated on 24/Dec/22

	Commented by mr W last updated on 24/Dec/22

	$\cos α = \frac{a - b}{a + b}$ $K F = K G = R$ $K D = a \sqrt{2 (1 - \cos α)} = \frac{2 a \sqrt{b}}{\sqrt{a + b}}$ $s i m i l a r l y K E = \frac{2 b \sqrt{a}}{\sqrt{a + b}}$ $D E = F D + G E = 2 R - D K - E K = 2 R - \frac{2 \sqrt{a b} (\sqrt{a} + \sqrt{b})}{\sqrt{a + b}}$ $D E = \sqrt{{(a + b)}^{2} - {(a - b)}^{2}} = 2 \sqrt{a b}$ $2 R - \frac{2 \sqrt{a b} (\sqrt{a} + \sqrt{b})}{\sqrt{a + b}} = 2 \sqrt{a b}$ $\Rightarrow R = \frac{\sqrt{a b} (\sqrt{a + b} + \sqrt{a} + \sqrt{b})}{\sqrt{a + b}}$
	Commented by ajfour last updated on 24/Dec/22

	$T h a n k s S i r . E x c e l l e n t w a y!$
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