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Question Number 76108 by Maclaurin Stickker last updated on 23/Dec/19

Commented by Maclaurin Stickker last updated on 23/Dec/19

$$inthefigure,thecircumferences$$$$haveradius8cmand6cmandthe$$$$distancebetweentheircentersis12cm.$$$$IfQP=PR,findQP.$$

Answered by mr W last updated on 23/Dec/19

Commented by mr W last updated on 23/Dec/19

$$r_1=8,r_2=6$$$$letAP=2b$$$$\sqrt{8^2-b^2}+\sqrt{6^2-b^2}=12$$$${\left(\sqrt{8^2-b^2}\right)}^2={(12-\sqrt{6^2-b^2})}^2$$$$0=116-24\sqrt{36-b^2}$$$$29=6\sqrt{36-b^2}$$$${\left(\frac{29}{6}\right)}^2=36-b^2$$$$\Rightarrow b=\sqrt{36-{\left(\frac{29}{6}\right)}^2}=\frac{\sqrt{455}}{6}$$$$AP=2b=\frac{\sqrt{455}}{3}$$$$\alpha ={\cos }^{-1}\frac{b}{r_1}={\cos }^{-1}\frac{\sqrt{455}}{48}$$$$\beta ={\cos }^{-1}\frac{b}{r_2}={\cos }^{-1}\frac{\sqrt{455}}{36}$$$$letQP=PR=x$$$$\sin \mathrm{\angle }QAP=\frac{x}{2r_1}=\frac{x}{16}$$$$\sin \mathrm{\angle }PAR=\frac{x}{2r_2}=\frac{x}{12}$$$$\mathrm{\angle }QAP+\mathrm{\angle }PAR=\pi -(\alpha +\beta )$$$$\cos (\mathrm{\angle }QAP+\mathrm{\angle }PAR)=-\cos (\alpha +\beta )$$$$\frac{\sqrt{({16}^2-x^2)({12}^2-x^2)}}{16\times 12}-\frac{x^2}{16\times 12}=\frac{43\times 29}{48\times 36}-\frac{455}{36\times 48}$$$$\sqrt{({16}^2-x^2)({12}^2-x^2)}-x^2=88$$$$({16}^2-x^2)({12}^2-x^2)={(88+x^2)}^2$$$${16}^2\times {12}^2-{88}^2=({16}^2+{12}^2+2\times 88)x^2$$$$455=9x^2$$$$\Rightarrow x=\frac{\sqrt{455}}{3}$$

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