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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

$$x=\sqrt{R^2-{\left(\frac{b}{2}\right)}^2}=\sqrt{R^2-\frac{b^2}{4}}$$$$y=\sqrt{R^2-\frac{a^2}{4}}$$$${(p-x)}^2+(p-y)={(R-p)}^2$$$$p^2+2(R-x-y)p-(R^2-x^2-y^2)=0$$$$\Rightarrow p=-R+x+y+\sqrt{2(R-x)(R-y)}$$$$similarly$$$$\Rightarrow q=-R-x-y+\sqrt{2(R+x)(R+y)}$$$$CD=\sqrt{2}(p+q)$$$$=-2\sqrt{2}R$$$$+2\sqrt{(R-\sqrt{R^2-\frac{a^2}{4}})(R-\sqrt{R^2-\frac{b^2}{4}})}$$$$+2\sqrt{(R+\sqrt{R^2-\frac{a^2}{4}})(R+\sqrt{R^2-\frac{b^2}{4}})}$$

Commented by ajfour last updated on 05/Nov/24

$$Thankyousir.$$$$verysimpledirecttreatment.$$

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