Question:
The equation \(\mathrm{Im}\left(\frac{iz - 2}{z - i}\right) + 1 = 0,\ z \in \mathbf{C},\ z \neq i\) represents a part of a circle having radius equal to :
The equation \(\mathrm{Im}\left(\frac{iz - 2}{z - i}\right) + 1 = 0,\ z \in \mathbf{C},\ z \neq i\) represents a part of a circle having radius equal to :
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Put z = x + iy, simplify the fraction and collect the imaginary part to get a circle equation.
Updated On: Nov 11, 2026
- 2
- 1
- \(\frac{3}{4}\)
- \(\frac{1}{2}\)
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The Correct Option is C
Solution and Explanation
Let \(z = x + iy\). Then \(iz - 2 = (-y-2) + ix\) and \(z - i = x + i(y-1)\).
\(\frac{iz-2}{z-i} = \frac{[(-y-2)+ix][x - i(y-1)]}{x^2 + (y-1)^2}\).
Imaginary part of the numerator \(= x\cdot x + (y+2)(y-1) = x^2 + y^2 + y - 2\).
So \(\mathrm{Im} = \frac{x^2+y^2+y-2}{x^2+(y-1)^2}\) and the condition \(\mathrm{Im} + 1 = 0\) gives \(x^2 + y^2 + y - 2 + x^2 + y^2 - 2y + 1 = 0\).
\(2x^2 + 2y^2 - y - 1 = 0 \Rightarrow x^2 + y^2 - \frac{y}{2} - \frac{1}{2} = 0\).
Radius \(= \sqrt{\frac{1}{16} + \frac{1}{2}} = \sqrt{\frac{9}{16}} = \frac{3}{4}\).
Answer: (3) \(\frac{3}{4}\)
\(\frac{iz-2}{z-i} = \frac{[(-y-2)+ix][x - i(y-1)]}{x^2 + (y-1)^2}\).
Imaginary part of the numerator \(= x\cdot x + (y+2)(y-1) = x^2 + y^2 + y - 2\).
So \(\mathrm{Im} = \frac{x^2+y^2+y-2}{x^2+(y-1)^2}\) and the condition \(\mathrm{Im} + 1 = 0\) gives \(x^2 + y^2 + y - 2 + x^2 + y^2 - 2y + 1 = 0\).
\(2x^2 + 2y^2 - y - 1 = 0 \Rightarrow x^2 + y^2 - \frac{y}{2} - \frac{1}{2} = 0\).
Radius \(= \sqrt{\frac{1}{16} + \frac{1}{2}} = \sqrt{\frac{9}{16}} = \frac{3}{4}\).
Answer: (3) \(\frac{3}{4}\)
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