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Question Number 72606 by TawaTawa last updated on 30/Oct/19

	Answered by mr W last updated on 30/Oct/19

	Commented by mr W last updated on 30/Oct/19

	${B C}^{2} = 12^{2} + {(2 r + 1)}^{2}$ $D B = 12$ $C D = \sqrt{{(r + 1)}^{2} - r^{2}} = \sqrt{2 r + 1}$ $12^{2} + {(2 r + 1)}^{2} = {(12 + \sqrt{2 r + 1})}^{2}$ ${(2 r + 1)}^{2} = 24 \sqrt{2 r + 1} + 2 r + 1$ $l e t u = \sqrt{2 r + 1}$ $u^{4} = 24 u + u^{2}$ $u^{3} - u - 24 = 0$ $(u - 3) (u^{2} + 3 u + 8) = 0$ $\Rightarrow u = 3$ $\Rightarrow \sqrt{2 r + 1} = 3$ $\Rightarrow r = 4$
	Commented by TawaTawa last updated on 30/Oct/19

	$God bless you sir .$
	Answered by Tanmay chaudhury last updated on 30/Oct/19

	$c i r c l e x^{2} + {(y - r)}^{2} = r^{2}$ $\frac{x}{12} + \frac{y}{2 r + 1} = 1 i s t a n g e n t \to B C$ $d i s t a n c e f r o m (0, r) t o B C i s r a d i u s$ $(2 r + 1) x + 12 y - 12 (2 r + 1) = 0$ $∣ \frac{(2 r + 1) \times 0 + 12 (r) - 24 r - 12}{\sqrt{{(2 r + 1)}^{2} + 12^{2}}} ∣= r$ $- 12 r - 12 = r \times \sqrt{4 r^{2} + 4 r + 1 + 144}$ $144 r^{2} + 144 + 288 r = 4 r^{4} + 4 r^{3} + 145 r^{2}$ $4 r^{4} + 4 r^{3} + r^{2} - 288 r - 144 = 0$ $r^{2} (4 r^{2} + 4 r + 1) - 144 (2 r + 1) = 0$ $r^{2} {(2 r + 1)}^{2} - 144 (2 r + 1) = 0$ $(2 r + 1) (2 r^{3} + r^{2} - 144) = 0$ $r \neq - 0.5$ $2 r^{3} + r^{2} - 128 - 16 = 0$ $2 (r^{3} - 64) + (r + 4) (r - 4) = 0$ $2 (r - 4) (r^{2} + 4 r + 16) + (r + 4) (r - 4) = 0$ $(r - 4) (2 r^{2} + 8 r + 32 + r + 4) = 0$ $s o r = 4 a n s w e r$
	Commented by TawaTawa last updated on 30/Oct/19

	$God bless you sir .$
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