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