Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
$\Rightarrow x=\lambda+3$
$\lambda=(x-3)\Rightarrow \lambda^2=(x-3)^2$
$y=-\sqrt{5-\lambda^2}\Rightarrow \lambda^2=5-y^2$
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App