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Question-213397




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 04/Nov/24
x=(√(R^2 −((b/2))^2 ))=(√(R^2 −(b^2 /4)))  y=(√(R^2 −(a^2 /4)))  (p−x)^2 +(p−y)=(R−p)^2   p^2 +2(R−x−y)p−(R^2 −x^2 −y^2 )=0  ⇒p=−R+x+y+(√(2(R−x)(R−y)))  similarly  ⇒q=−R−x−y+(√(2(R+x)(R+y)))  CD=(√2)(p+q)          =−2(√2)R           +2(√((R−(√(R^2 −(a^2 /4))))(R−(√(R^2 −(b^2 /4))))))          +2(√((R+(√(R^2 −(a^2 /4))))(R+(√(R^2 −(b^2 /4))))))
$${x}=\sqrt{{R}^{\mathrm{2}} −\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${y}=\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\left({p}−{x}\right)^{\mathrm{2}} +\left({p}−{y}\right)=\left({R}−{p}\right)^{\mathrm{2}} \\ $$$${p}^{\mathrm{2}} +\mathrm{2}\left({R}−{x}−{y}\right){p}−\left({R}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{p}=−{R}+{x}+{y}+\sqrt{\mathrm{2}\left({R}−{x}\right)\left({R}−{y}\right)} \\ $$$${similarly} \\ $$$$\Rightarrow{q}=−{R}−{x}−{y}+\sqrt{\mathrm{2}\left({R}+{x}\right)\left({R}+{y}\right)} \\ $$$${CD}=\sqrt{\mathrm{2}}\left({p}+{q}\right) \\ $$$$\:\:\:\:\:\:\:\:=−\mathrm{2}\sqrt{\mathrm{2}}{R} \\ $$$$\:\:\:\:\:\:\:\:\:+\mathrm{2}\sqrt{\left({R}−\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\right)\left({R}−\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}\right)} \\ $$$$\:\:\:\:\:\:\:\:+\mathrm{2}\sqrt{\left({R}+\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\right)\left({R}+\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}\right)} \\ $$
Commented by ajfour last updated on 04/Nov/24
Thank you sir. CD is indeed lengthy.  very simple direct treatment .
$${Thank}\:{you}\:{sir}.\:{CD}\:{is}\:{indeed}\:{lengthy}. \\ $$$${very}\:{simple}\:{direct}\:{treatment}\:. \\ $$

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