Question

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

• A. $x^2+y^2-2x-2y+1+{\sin }^2{72}^o=0$
• B. $x^2+y^2-2x-2y+{\cos }^2{72}^o=0$
• C. $x^2+y^2-2x-2y+1+{\cos }^2{18}^o=0$
  
• D. $x^2+y^2-2x-2y+{\sin }^2{72}^o=0$

Answer 1

Answer: (A), (C)

The correct options are
A $x^2+y^2-2x-2y+1+{\sin }^2{72}^o=0$
C $x^2+y^2-2x-2y+1+{\cos }^2{18}^o=0$

Diagonal intersects at $(1,1)$

Length of the diagonal $=\sin 36°$

Diagonals intersect at $y-2x+1=0$ and $2y-x-1=0$

Length of the diagonal $\sqrt{a^2+a^2-2a^2\cos 108°}$

$=\sqrt{2}a\sqrt{1-\cos 108°}$

$=2+\sin 36°$

$\tan 36°=2+\sin 36°$

$r=\cos 72°$

${(x-1)}^2+{(y-1)}^2={\cos }^{2}72°$

$x^2+y^2-2x-2y+1+{\sin }^{2}72°=0$

and $x^2+y^2-2x-2y+1+{\cos }^{2}18°=0$