If `z =(lambda+3)+isqrt((5-lambda^2))` ; then the locus of ` z` is

Options

A. ellispe
B. semicircle
C. parabola
D. none of these

Answer

B

Explanation

Let `z = x + iy`. Then,
`x=lambda+3and y=-sqrt(5-lamba^(2))`
`rArr (x-3)^(2)=lambda^(2)" " (1)`
and `y^(2)=5-lambda^(2)" "(2)`
From Eqs. (1) and (2),
`(x-3)^(2)=5-y^(2)or(x-3)^(2)+y^(2)=5`
Obviously it is a semicircle as `ylt0`. Hence, part of the circle lies below the x-axis.