Question

Statement 1: Consider two curves $C_1:\overline{z}+iz-i\overline{z}+b=0$ and $C_2:z\bar{z}+(1-i)z+(1+i)\bar{z}-4=0$ where $(b\in R,z=x+$ iy and $i=\sqrt{-1})$. If $C_1$ and $C_2$ intersects orthogonally then $b=-2$.
Statement 2: If two curves intersects orthogonally then the angle between the tangents at all their points of intersection is $\frac{\pi }{2}$.

• A. Statement- 1 is true, statement- 2 is true and statement- 2 is correct explanation for statement- 1 .
• B. Statement- 1 is true, statement- 2 is true and statement- 2 is NOT the correct explanation for statement- 1 .
• C. Statement- 1 is true, statement- 2 is false.
• D. Statement- 1 is false, statement- 2 is true.

Answer 1

Answer: (D)

$C_1:z\bar{z}+iz-i\bar{z}+b=0$
Put $z=x+iy$
$x^2+y^2+i(x+iy)-i(x-iy)+b=0$
$\Rightarrow x^2+y^2-2y+b=0$
Also, $\bar{z}\bar{z}+(1-i)z+(1+i)\bar{z}-4=0$
$\Rightarrow x^2+y^2+(1-i)(x+iy)+(1+i)(x-iy)-4=0$.
$\Rightarrow x^2+y^2+2x+2y-4=0$
$\therefore$ Using condition of orthogonally, we get
$2(0\times 1+(-1)(1))=b-4\Rightarrow -2=b-4\Rightarrow b=2.$
$\therefore$ S-1 is false, but S-2 is true.