A) Circle whose centre is at \[\left( -\frac{1}{2},-\frac{3}{2} \right)\]
B) Straight line whose slope is \[-\frac{2}{3}\]
C) Circle whose diameter is \[\frac{\sqrt{5}}{2}\]
D) Straight line whose slope is \[\frac{3}{2}\]
Correct Answer: C
Solution :
| [c] |
| \[\left( \frac{z-1}{2z+i} \right)=\left( \frac{x-1+iy}{2x+i(1+2y)} \right)\] |
| \[=\frac{(x-1+iy)(2x-i)(1+2y)}{4{{x}^{2}}+{{(2y+1)}^{2}}}\] |
| As its real part is 1 |
| \[\Rightarrow \frac{2{{x}^{2}}-2x+y+2{{y}^{2}}}{4{{x}^{2}}+4{{y}^{2}}+1+4y}=1\] |
| \[\Rightarrow 2{{x}^{2}}+2{{y}^{2}}+2x+3y+1=0\] |
| i.e. \[{{x}^{2}}+{{y}^{2}}+x+\frac{3}{2}y+\frac{1}{2}=0\] |
| \[\Rightarrow {{\left( x+\frac{1}{2} \right)}^{2}}+{{\left( y+\frac{3}{4} \right)}^{2}}=\frac{5}{16}\] |
| i.e. circle with diameter \[\frac{\sqrt{5}}{2}\] |
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