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If $z=(\lambda+3)-i\sqrt{5-\lambda^2}$,then locus of $z$ is

$\begin{array}{1 1}(A)\;\text{Circle}&(B)\;\text{Semicircle}\\(C)\;\text{Quatercircle}&(D)\;\text{Straight line}\end{array} $

1 Answer

$z=x+iy=(\lambda+3)-i\sqrt{5-\lambda^2}$
$\Rightarrow x=\lambda+3$
$y=-\sqrt{5-\lambda^2}$
$\lambda=(x-3)\Rightarrow \lambda^2=(x-3)^2$
$y=-\sqrt{5-\lambda^2}\Rightarrow \lambda^2=5-y^2$
Comparing $(x-3)^2=5-y^2$
$\Rightarrow (x-3)^2+y^2=5$
$\Rightarrow $Equation of circle
But $y=-\sqrt{5-\lambda^2}$ (i.e) y=-ve
Thus locus is not full circle but only half circle with -ve y
Hence (B) is the correct answer.
answered Apr 10, 2014 by sreemathi.v
 
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