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  • If the equation of any two diagonals of a regular pentagon belongs to family of lines<img alt="\mathrm{(1+2 \lambda) y-(2+\lambda) x+1-\lambda=0}" src="https://entrancecorner.oncodecogs.com/gif.lat

If the equation of any two diagonals of a regular pentagon belongs to family of lines\mathrm{(1+2 \lambda) y-(2+\lambda) x+1-\lambda=0} and their lengths are sin \mathrm{36^{\circ}}, then locus of centre of circle circumscribing the given pentagon (the triangles formed by these diagonals with sides of pentagon have no side common), is

Option: 1

\mathrm{x^2+y^2-2 x-2 y+1+\sin ^2 72^{\circ}=0}


Option: 2

\mathrm{x^2+y^2-2 x-2 y+\cos ^2 72^0=0}


Option: 3

\mathrm{x^2+y^2-2 x-2 y+1+\cos ^2 72^0=0}


Option: 4

\mathrm{x^2+y^2-2 x-2 y+\sin ^2 72^0=0}


Answers (1)

best_answer

Point of intersection of diagonals lie on circmucircle i.e. (1, 1)

\mathrm{\begin{aligned} & I=2 R \sin 72^{\circ} \\ & R=\frac{\sin 36^{\circ}}{2 \sin 72^{\circ}}=\cos 72^{\circ} \\ & \Rightarrow \text { locus is }(x-1)^2+(y-1)^2=\cos ^2 72^{\circ} \\ & \Rightarrow x^2+y^2-2 x-2 y+1+\sin ^2 72^{\circ}=0 \end{aligned}}

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Sanket Gandhi

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