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Question Number 170073 by mr W last updated on 15/May/22

Commented by mr W last updated on 15/May/22

$F i n d ∣ C D ∣_{m i n} = ?, ∣ C D ∣_{m a x} = ?$
Commented by cortano1 last updated on 16/May/22

$m a x = 5 - \frac{4 \sqrt{3}}{3} = \frac{1}{3} (15 - 4 \sqrt{3}) ?$
Commented by mr W last updated on 16/May/22

$n o .$
	Answered by ajfour last updated on 16/May/22

	Commented by ajfour last updated on 16/May/22

	$c e n t r e o f c i r c l e P (n o t A)$ $P (\frac{2}{\sqrt{3}}, 2) [B o r i g i n]$ $\cos 30 ° = \frac{2}{r} \Rightarrow r = \frac{4}{\sqrt{3}}$ ${C P}^{2} = {(5 - \frac{2 \sqrt{3}}{3})}^{2} + 2^{2}$ $= 29 + \frac{4}{3} - \frac{20 \sqrt{3}}{3}$ $= \frac{91 - 20 \sqrt{3}}{3}$ ${C D}_{m i n} = C P - r$
	Commented by mr W last updated on 16/May/22

	$t h a n k s s i r!$
	Commented by ajfour last updated on 16/May/22

	$l e t m e t h i n k . .$
	Commented by Tawa11 last updated on 08/Oct/22

	$Great sir$
	Answered by mr W last updated on 16/May/22

	Commented by ajfour last updated on 16/May/22

	$y e a h, s u r e g u e s s m n o t f o c u s s i n g . .$
	Commented by mr W last updated on 16/May/22

	$r = \frac{2}{\cos 30 °} = \frac{4 \sqrt{3}}{3}$ $C E = \sqrt{{(5 - \frac{2}{\sqrt{3}})}^{2} + 2^{2}} = \frac{\sqrt{273 - 60 \sqrt{3}}}{3}$ ${C E}^{'} = \sqrt{{(5 + \frac{2}{\sqrt{3}})}^{2} + 2^{2}} = \frac{\sqrt{273 + 60 \sqrt{3}}}{3}$ ${C D}_{m i n} = C E - r = \frac{\sqrt{273 - 60 \sqrt{3}} - 4 \sqrt{3}}{3}$ ${C D}_{m a x} = {C E}^{'} + r = \frac{\sqrt{273 + 60 \sqrt{3}} + 4 \sqrt{3}}{3}$
	Commented by ajfour last updated on 16/May/22

	$f a b u l o u s .$
	Answered by cortano1 last updated on 17/May/22

	Commented by mr W last updated on 17/May/22

	$g o o d!$
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