If the equation of any two diagonals of a regular pentagon belongs to the family of lines (1+2 λ) y -(2+λ) x +1-λ= 0 and their lengths are sin 36°, 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

Question

If the equation of any two diagonals of a regular pentagon belongs to the family of lines (1+2 λ) y -(2+λ) x +1-λ= 0 and their lengths are sin 36°, 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) x2+y2-2 x-2 y+1+ sin 2 72°=0 (B) x2+y2-2 x-2 y+ cos 2 72°=0 (C) x2+y2-2 x-2 y+1+ cos 2 72°=0 (D) x2+y2-2 x-2 y+ sin 2 72°=0

Tardigrade Admin · Accepted Answer

The point of intersection of diagonals, i.e.,

(1, 1)

, lies on the circumcircle.

Hence,

I = 2 R sin 7 2^{\circ}

R = \frac{s i n 3 6 ^{\circ}}{2 s i n 7 2 ^{\circ}} = cos 7 2^{\circ}

Therefore, the locus is

(x - 1)^{2} + (y - 1)^{2} = cos^{2} 7 2^{\circ}

.

x^{2} + y^{2} - 2 x - 2 y + 1 + sin^{2} 7 2^{\circ} = 0

$x^{2} + y^{2} - 2 x - 2 y + 1 + sin^{2} 7 2^{\circ} = 0$

$x^{2} + y^{2} - 2 x - 2 y + cos^{2} 7 2^{\circ} = 0$

$x^{2} + y^{2} - 2 x - 2 y + 1 + cos^{2} 7 2^{\circ} = 0$

$x^{2} + y^{2} - 2 x - 2 y + sin^{2} 7 2^{\circ} = 0$

x2+y2−2x−2y+1+sin272∘=0

x2+y2−2x−2y+cos272∘=0

x2+y2−2x−2y+1+cos272∘=0

x2+y2−2x−2y+sin272∘=0

The point of intersection of diagonals, i.e., (1,1), lies on the circumcircle. Hence, I=2Rsin72∘ R=2sin72∘sin36∘​=cos72∘ Therefore, the locus is (x−1)2+(y−1)2=cos272∘. x2+y2−2x−2y+1+sin272∘=0

$x^{2} + y^{2} - 2 x - 2 y + 1 + sin^{2} 7 2^{\circ} = 0$

$x^{2} + y^{2} - 2 x - 2 y + cos^{2} 7 2^{\circ} = 0$

$x^{2} + y^{2} - 2 x - 2 y + 1 + cos^{2} 7 2^{\circ} = 0$

$x^{2} + y^{2} - 2 x - 2 y + sin^{2} 7 2^{\circ} = 0$