The equation (1+i)z−2(1+i)z+4=k does not represent a circle when k is
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a
2
b
π
c
e
d
1
answer is D.
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Detailed Solution
We can write the equation as z−2(1+i)z+4(1+i)=k (1)But 21+i=12−i21+i=1−i and 41+i=2(1−i) z−(1−i)z+2(1−i)=k (2)This will not represent a circle if k = 1. When k = 1, (2) represents perpendicular bisector of the segment joining –2(1 – i) and 1 – i.