1. Interpret the set

Given $S=\left\{z\in \mathbb C: z=\frac{1}{a+ibt},\ t\in \mathbb R,\ t\neq 0\right\},$ where $a,b\in \mathbb R$ and $a^2+b^2\neq 0$.

Let $z=x+iy=\frac{1}{a+ibt}.$ We must find the locus of $(x,y)$.

2. Write $z$ in Cartesian form

Rationalizing, $z=\frac{1}{a+ibt}\cdot \frac{a-ibt}{a-ibt}=\frac{a-ibt}{a^2+b^2t^2}.$ Hence $x=\frac{a}{a^2+b^2t^2},\qquad y=\frac{-bt}{a^2+b^2t^2}.$

3. Eliminate the parameter $t$ when $a\neq 0$ and $b\neq 0$

We compute $x^2+y^2=\frac{a^2+b^2t^2}{(a^2+b^2t^2)^2}=\frac{1}{a^2+b^2t^2}.$ Therefore, $x=a(x^2+y^2).$ Rearranging, $x^2+y^2-\frac{x}{a}=0.$ Complete the square: $x^2-\frac{x}{a}+y^2=0$ $\left(x-\frac{1}{2a}\right)^2+y^2=\frac{1}{4a^2}.$

So for $a\neq 0, b\neq 0$, the locus is the circle with centre $\left(\frac{1}{2a},0\right)$ and radius $\frac{1}{2|a|}.$

Now compare with the options:

• If $a>0$, radius $\frac{1}{2a}$ and centre $\left(\frac{1}{2a},0\right)$: Option A is correct.
• If $a<0$, the same circle can be written as centre $\left(-\frac{1}{2|a|},0\right)=\left(\frac{1}{2a},0\right)$ and radius $\frac{1}{2|a|}=-\frac{1}{2a}$. Thus Option B is also correct.

So the geometric statement for $a<0$ given in B is valid.

4. Case $a\neq 0,\ b=0$

Then $z=\frac{1}{a},$ which is a fixed real number. Hence $(x,y)=(1/a,0)$, which lies on the $x$-axis.

So Option C is correct.

5. Case $a=0,\ b\neq 0$

Then $z=\frac{1}{ibt}=\frac{-i}{bt}.$ Thus $x=0,\qquad y=-\frac{1}{bt}.$ Hence $(x,y)$ lies on the $y$-axis.

So Option D is correct.

6. Evaluate all options

• A: Correct for $a>0,\ b\neq 0$.
• B: Correct for $a<0,\ b\neq 0$.
• C: Correct for $a\neq 0,\ b=0$.
• D: Correct for $a=0,\ b\neq 0$.

Thus the mathematically correct set of options is $\boxed{A,B,C,D}.$

7. Compare with stored correct answer

Stored answer: $D, C, A$.

But option B is also correct, because when $a<0$, $\left(x-\frac{1}{2a}\right)^2+y^2=\frac{1}{4a^2}$ represents a circle with centre $\left(-\frac{1}{2|a|},0\right)$ and radius $\frac{1}{2|a|}=-\frac{1}{2a}$, exactly matching option B.

So I disagree with the stored answer.