Question

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

• A. $x^{2}+y^{2}-2 x-2 y+1+\sin ^{2} 72^{\circ}=0$
• B. $x^{2}+y^{2}-2 x-2 y+\cos ^{2} 72^{\circ}=0$
• C. $x^{2}+y^{2}-2 x-2 y+1+\cos ^{2} 72^{\circ}=0$
• D. $x^{2}+y^{2}-2 x-2 y+\sin ^{2} 72^{\circ}=0$

Answer 1

Answer: (A)

The point of intersection of diagonals, i.e., $(1,1)$, lies on the circumcircle.

Hence,
$I =2 R \sin 72^{\circ} $
$R =\frac{\sin 36^{\circ}}{2 \sin 72^{\circ}}=\cos 72^{\circ}$
Therefore, the locus is $(x-1)^{2}+(y-1)^{2}=\cos ^{2} 72^{\circ}$.
$x^{2}+y^{2}-2 x-2 y+1+\sin ^{2} 72^{\circ}=0$