Question Number 213397 by ajfour last updated on 04/Nov/24
Answered by mr W last updated on 04/Nov/24
Commented by mr W last updated on 05/Nov/24
$${x}=\sqrt{{R}^{\mathrm{2}} −\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${y}=\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\left({p}−{x}\right)^{\mathrm{2}} +\left({p}−{y}\right)=\left({R}−{p}\right)^{\mathrm{2}} \\ $$$${p}^{\mathrm{2}} +\mathrm{2}\left({R}−{x}−{y}\right){p}−\left({R}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{p}=−{R}+{x}+{y}+\sqrt{\mathrm{2}\left({R}−{x}\right)\left({R}−{y}\right)} \\ $$$${similarly} \\ $$$$\Rightarrow{q}=−{R}−{x}−{y}+\sqrt{\mathrm{2}\left({R}+{x}\right)\left({R}+{y}\right)} \\ $$$${CD}=\sqrt{\mathrm{2}}\left({p}+{q}\right) \\ $$$$\:\:\:=−\mathrm{2}\sqrt{\mathrm{2}}{R} \\ $$$$\:\:\:+\mathrm{2}\sqrt{\left({R}−\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\right)\left({R}−\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}\right)} \\ $$$$\:\:\:+\mathrm{2}\sqrt{\left({R}+\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\right)\left({R}+\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}\right)} \\ $$
Commented by ajfour last updated on 05/Nov/24
$${Thank}\:{you}\:{sir}. \\ $$$${very}\:{simple}\:{direct}\:{treatment}. \\ $$