the equation of circle is x^2+y^2=4. a regular hexagon is inscribed in the circle whose vertex is (2,0). then a consecutive vertex has the coordinates ____

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$$\begin{gathered} {\mathrm{x}}^2+{\mathrm{y}}^2=4=2^2 \\ \mathrm{Hence} \mathrm{circumradius}=\mathrm{R}=2 \\ \mathrm{If} \mathrm{side} \mathrm{of} \mathrm{hexagon} \mathrm{is} \mathrm{a}, \mathrm{then} \mathrm{we} \mathrm{know} \mathrm{that} \\ \mathrm{a}=2\mathrm{Rsin}\frac{\mathrm{\pi }}{6} \\ \Rightarrow \mathrm{a}=2\times 2\times \frac{1}{2}=2 \\ \mathrm{Hence} \mathrm{we} \mathrm{need} \mathrm{a} \mathrm{point} \mathrm{on} \mathrm{circle} \mathrm{which} \mathrm{is} 2 \mathrm{unit} \mathrm{away} \mathrm{from} \mathrm{Q}\left(2,0\right) \\ \mathrm{Let} \mathrm{that} \mathrm{point} \mathrm{be} \mathrm{P}\left(2\mathrm{cos\theta },2\mathrm{sin\theta }\right) \\ \mathrm{Hence} \\ \mathrm{PQ}=2 \\ \Rightarrow \sqrt{{\left(2\mathrm{cos\theta }-2\right)}^2+{\left(2\mathrm{sin\theta }\right)}^2}=2 \\ \Rightarrow {\left(2\mathrm{cos\theta }-2\right)}^2+{\left(2\mathrm{sin\theta }\right)}^2=4 \\ \Rightarrow 4{\left(\mathrm{cos\theta }-1\right)}^2+4{\sin }^2\mathrm{\theta }=4 \\ \Rightarrow {\left(\mathrm{cos\theta }-1\right)}^2+{\sin }^2\mathrm{\theta }=1 \\ \Rightarrow {\cos }^2\mathrm{\theta }+1-2\mathrm{cos\theta }+{\sin }^2\mathrm{\theta }=1 \\ \Rightarrow {\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }-2\mathrm{cos\theta }=0 \\ \Rightarrow 1=2\mathrm{cos\theta } \\ \Rightarrow \mathrm{cos\theta }=\frac{1}{2} \\ \mathrm{Hence} \\ \mathrm{\theta }=\frac{\mathrm{\pi }}{3} \mathrm{or} \frac{5\mathrm{\pi }}{3} \\ \left(2\mathrm{cos\theta },2\mathrm{sin\theta }\right)=\left(2\cos \frac{\mathrm{\pi }}{3},2\sin \frac{\mathrm{\pi }}{3}\right)=\left(2\times \frac{1}{2},2\times \frac{\sqrt{3}}{2}\right)\mathbf{=}\left(\mathbf{1}\mathbf{,}\sqrt{\mathbf{3}}\right) \\ \mathrm{or} \\ \left(2\mathrm{cos\theta },2\mathrm{sin\theta }\right)=\left(2\cos \frac{5\mathrm{\pi }}{3},2\sin \frac{5\mathrm{\pi }}{3}\right)=\left(2\times \frac{1}{2},-2\times \frac{\sqrt{3}}{2}\right)\mathbf{=}\left(\mathbf{1}\mathbf{,}\mathbf{-}\sqrt{\mathbf{3}}\right) \end{gathered}$$

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