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Question Number 42538 by ajfour last updated on 27/Aug/18

Commented by ajfour last updated on 29/Aug/18

$Q . 42446 F i n d r_{m a x} i f r a d i u s o f$ $l a r g e r c i r c l e i s R .$
	Answered by ajfour last updated on 29/Aug/18

	$E q . o f p a r a b o l a$ $y = {a x}^{2} \Rightarrow \frac{d y}{d x} ∣_{(x_{1}, y_{1})} = 2 {a x}_{1}$ $e q . o f t a n g e n t t o p a r a b o l a a t$ $(x_{1}, {a x}_{1}^{2}) i s$ $y - {a x}_{1}^{2} = 2 {a x}_{1} (x - x_{1})$ $s l o p e o f c o m m o n t a n g e n t t o$ $c i r c l e s a t T : m = \cot θ$ $T (R \cos θ, R - R \sin θ)$ $l e t 2 {a x}_{1} = \cot θ$ $y_{1} = {a x}_{1}^{2}$ $y_{1} = {a x}_{1}^{2} a n d \cot θ = 2 {a x}_{1}$ $x_{1} = (R - 2 r) \cos θ$ $y_{1} = R - (R - 2 r) \sin θ$ $\Rightarrow R - (R - 2 r) \sin θ = {a x}_{1}^{2}$ $R - \sqrt{{(R - 2 r)}^{2} - x_{1}^{2}} = {a x}_{1}^{2}$ $\Rightarrow {(R - 2 r)}^{2} - x_{1}^{2} = a^{2} x_{1}^{4} + R^{2} - 2 {a R x}_{1}^{2}$ $o n l y o n e v a l u e o f x_{1} \Rightarrow$ ${(1 - 2 a R)}^{2} = 4 a^{2} (4 r R - 4 r^{2})$ $\Rightarrow r^{2} - r R + \frac{{(1 - 2 a R)}^{2}}{16 a^{2}} = 0$ $r = \frac{R}{2} - \sqrt{\frac{R^{2}}{4} - \frac{(1 - 4 a R + 4 a^{2} R^{2})}{16 a^{2}}}$ $\Rightarrow r = \frac{R}{2} - \sqrt{\frac{R}{4 a} - \frac{1}{16 a^{2}}}$ $\Rightarrow r = \frac{R}{2} - \frac{\sqrt{4 a R - 1}}{4 a} .$
	Commented by ajfour last updated on 29/Aug/18

	$o n e o r t w o o t h e r r o u t e s s t i l l$ $y i e l d d i f f e r e n t a n s w e r s . .!$ $p l e a s e d i s c u s s a n d h e l p . .$
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