Question

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

• A. circle
• B. parabola
• C. straight line
• D. None of these

Answer 1

The correct option is A circle

Given: $z=\lambda +3+i\sqrt{(3-{\lambda }^2)}$

Comparing with $Z=x+iy$

Simplifying we get

$x=\lambda +3$ ------(i)

$\Rightarrow y=\sqrt{3-{\lambda }^2}$

$\Rightarrow y^2=3-{\lambda }^2$

$\Rightarrow {\lambda }^2=3-y^2$-----(ii)

Hence

From (i) we can say that

$\Rightarrow (x-3)=\lambda$

Squaring both the side

$\Rightarrow (x-3{)}^2=3-y^2$-----from (ii)

$\Rightarrow (x-3{)}^2+y^2=3$

Hence $z$ lies on a circle.