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Question Number 137374 by oooooooo last updated on 02/Apr/21

Answered by mr W last updated on 02/Apr/21

$$letOA=OD=R=radius$$$${(\frac{R}{\sqrt{2}}-4)}^2+{(\frac{R}{\sqrt{2}}+3)}^2=R^2...\left(i\right)$$$${(\frac{R}{\sqrt{2}}-x)}^2+{(\frac{R}{\sqrt{2}}+5)}^2=R^2...\left(ii\right)$$$$\left(i\right):$$$$-\sqrt{2}R+25=0\Rightarrow R=\frac{25}{\sqrt{2}}$$$$\left(ii\right):$$$${(\frac{25}{2}-x)}^2+{(\frac{25}{2}+5)}^2=\frac{{25}^2}{2}$$$${(\frac{25}{2}-x)}^2={\left(\frac{5}{2}\right)}^2$$$$\frac{25}{2}-x=\pm \frac{5}{2}$$$$x=\frac{25}{2}\pm \frac{5}{2}=15or10$$

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