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Question-206079




Question Number 206079 by mr W last updated on 06/Apr/24
Answered by A5T last updated on 06/Apr/24
Commented by A5T last updated on 06/Apr/24
x is not dependent on the position of A on the  diameter, otherwise, x would not be uniquely  defined.  So,consider the case when A is coincident with  the center, then length x subtends an angle of  60° at the centre, which makes it equal to the  radius⇒x=8. This must also be true for all  positions of A.
$${x}\:{is}\:{not}\:{dependent}\:{on}\:{the}\:{position}\:{of}\:{A}\:{on}\:{the} \\ $$$${diameter},\:{otherwise},\:{x}\:{would}\:{not}\:{be}\:{uniquely} \\ $$$${defined}. \\ $$$${So},{consider}\:{the}\:{case}\:{when}\:{A}\:{is}\:{coincident}\:{with} \\ $$$${the}\:{center},\:{then}\:{length}\:{x}\:{subtends}\:{an}\:{angle}\:{of} \\ $$$$\mathrm{60}°\:{at}\:{the}\:{centre},\:{which}\:{makes}\:{it}\:{equal}\:{to}\:{the} \\ $$$${radius}\Rightarrow{x}=\mathrm{8}.\:{This}\:{must}\:{also}\:{be}\:{true}\:{for}\:{all} \\ $$$${positions}\:{of}\:{A}. \\ $$
Answered by mr W last updated on 06/Apr/24
Commented by mr W last updated on 06/Apr/24
x is a chord corresponding to a  central angle of 60°.   ⇒x=R=8
$${x}\:{is}\:{a}\:{chord}\:{corresponding}\:{to}\:{a} \\ $$$${central}\:{angle}\:{of}\:\mathrm{60}°.\: \\ $$$$\Rightarrow{x}={R}=\mathrm{8} \\ $$
Answered by mr W last updated on 06/Apr/24
Commented by mr W last updated on 06/Apr/24
R^2 =p^2 +a^2 +pa  R^2 =q^2 +a^2 −qa  ⇒p, −q are roots of z^2 +az+a^2 −R^2 =0  ⇒p−q=−a, −pq=a^2 −R^2   x^2 =p^2 +q^2 −pq=(p−q)^2 +pq      =(−a)^2 −(a^2 −R^2 )=R^2   ⇒x=R=8 ✓
$${R}^{\mathrm{2}} ={p}^{\mathrm{2}} +{a}^{\mathrm{2}} +{pa} \\ $$$${R}^{\mathrm{2}} ={q}^{\mathrm{2}} +{a}^{\mathrm{2}} −{qa} \\ $$$$\Rightarrow{p},\:−{q}\:{are}\:{roots}\:{of}\:{z}^{\mathrm{2}} +{az}+{a}^{\mathrm{2}} −{R}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{p}−{q}=−{a},\:−{pq}={a}^{\mathrm{2}} −{R}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} −{pq}=\left({p}−{q}\right)^{\mathrm{2}} +{pq} \\ $$$$\:\:\:\:=\left(−{a}\right)^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)={R}^{\mathrm{2}} \\ $$$$\Rightarrow{x}={R}=\mathrm{8}\:\checkmark \\ $$

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