complex numbers (locus) doubt

\hspace{-16} $Here $\bf{\mid z \mid = 2}$ and Here We have to find the value of $\bf{(a+bz)}$\\\\\\ $\bf{\mid a+bz\mid \leq \mid a \mid+\mid b \mid .\mid z \mid\Leftrightarrow \mid a+bz\mid \leq a+2b\;\;.\; a,b\in \mathbb{R}}$\\\\\\ $\bf{\mid a+bz \mid^2 \leq (a+2b)^2}$\\\\\\ Let $\bf{z=x+iy}$. Then $\bf{\mid a+b(x+iy)\mid^2 \leq (a+2b)^2}$\\\\\\ $\bf{(a+bx)^2+(by)^2\leq a^2+4b^2+4ab}$\\\\\\ $\bf{a^2+b^2x^2+2abx+b^2y^2\leq a^2+4b^2+4ab}$\\\\\\ $\bf{b^2x^2+b^2y^2+2abx-4b^2-4ab\leq 0}$\\\\\\ $\bf{x^2+y^2+2.\frac{a}{b}.x-4-4.\frac{a}{b}\leq 0}$\\\\\\ $\bf{\left(x+\frac{a}{b}\right)^2+y^2\leq \left(\frac{a}{b}\right)^2+\frac{4a}{b}+4}$\\\\\\ $\bf{\left(x+\frac{a}{b}\right)^2+y^2\leq \left(\frac{a}{b}+2\right)^2}$

yes, because its easily seen that |t-a| = |b||z| = 2|b|

Yes Bhatt Sir ,

Yes Rahul You are Right.

Locus is Simply a Curve not a closed Figure(My Mistake)

|a+bz|=b|ab+z|

if |z|=2
the |ab+z|=2

centre of circle shifted from 0,0 to (-ab,0)

thus
|a+bz|=2b
thus

its a circle