The equation Im(iz-2)/(z-i)+1=0, z∈C, z≠i represents a part of a circle having radius equal to :

Correct answer: (C) 3/4. Let z=x+iy, then Im (ix-y-2)/(x+y-1i) (x-y-1i)/(x-y-1i)+1=(y-1 y+2+x^2)/(x^2+y-1^2)+1=0 ⇒2x^2+2y^2-y-1=0 ⇒x^2+y^2-y/2-1/2=0 Center (0, 1/4) Radius =√(g^2+f^2-c) Radius= √(116+12)= √916=3/4

Q33 / 90 · JEE Main 2017 (09 Apr Online) · Mathematics

The equation $I m (\frac{i z - 2}{z - i}) + 1 = 0, z \in C, z \neq i$ represents a part of a circle having radius equal to :

A
$1$
B
$2$
Correct answer
C
$\frac{3}{4}$
D
$\frac{1}{2}$

Solution

Let $z = x + i y,$ then

$I m [(\frac{i x - y - 2}{x + (y - 1) i}) (\frac{x - (y - 1) i}{x - (y - 1) i})] + 1 = \frac{(y - 1) (y + 2) + x^{2}}{x^{2} + {(y - 1)}^{2}} + 1 = 0$

$\Rightarrow 2 x^{2} + 2 y^{2} - y - 1 = 0$

$\Rightarrow x^{2} + y^{2} - \frac{y}{2} - \frac{1}{2} = 0$

Center $(0, \frac{1}{4})$

Radius $= \sqrt{g^{2} + f^{2} - c}$

Radius $= \sqrt{\frac{1}{16} + \frac{1}{2}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$

Hence, the correct answer is option: 'C'.

The equation Imiz-2z-i+1=0, z∈C, z≠i represents a part of a circle having radius equal to :

Solution

Relevant Questions

The equation $I m (\frac{i z - 2}{z - i}) + 1 = 0, z \in C, z \neq i$ represents a part of a circle having radius equal to :