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Question Number 185800 by Rupesh123 last updated on 27/Jan/23

	Answered by ajfour last updated on 27/Jan/23

	$l e t c e n t e r o f s m a l l c i r c l e (- h, - k)$ $h e n c e e q . {(x + h)}^{2} + {(y + k)}^{2} = r^{2}$ $s i d e o f e q l . t r i a n g l e a = R \sqrt{3}$ $l e n g t h o f t a n g e n t f r o m A (0, R)$ $i s A P = \sqrt{h^{2} + {(R + k)}^{2} - r^{2}} . . (i)$ $b u t k = \frac{R}{2} + r . . . (i i)$ $a n d h^{2} + k^{2} = {(R - r)}^{2} . . . (i i i)$ $u s i n g (i i) a n d (i i i) i n (i)$ ${A P}^{2} = {(R - r)}^{2} + R^{2} - r^{2} + 2 R (\frac{R}{2} + r)$ $\Rightarrow A P^{2} = 3 R^{2}$ $A P = R \sqrt{3} = B C$
	Commented by mr W last updated on 28/Jan/23

	$n i c e a p p r o a c h!$
	Answered by mr W last updated on 28/Jan/23

	Commented by mr W last updated on 28/Jan/23

	$R = r a d i u s o f b i g c i r c l e$ $r = r a d i u s o f s m a l l c i r c l e$ $O Q = \frac{R}{2}$ $B C = \sqrt{3} R$ $({A S}^{2} =) {A P}^{2} + r^{2} = {(R + \frac{R}{2} + r)}^{2} + {(R - r)}^{2} - {(\frac{R}{2} + r)}^{2}$ ${A P}^{2} = 3 R^{2}$ $\Rightarrow A P = \sqrt{3} R = B C ✓$
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