Let the centre by $(h,k)$ and radius be $k$
Then $\sqrt {h^2+(k-1)^2}=1+k$ $$h^2=4k$$
I checked the graph of the circle and locus, and got this
If the blue curve is the centre of the new, there has to be some point $(x,y)$ beyond which the condition isn’t satisfied. This is the right answer, but there are two options, with another condition
A)$ \{(x,y): x^2=4y\} \cup \{(x,y): y\le 0\}$
B) $ \{(x,y): x^2=4y\} \cup \{(0,y): y\le 0\}$
Which is the correct option
To show that the option A is incorrect, we have to show there exist no circunference such that they are tangent to $x^2+(y-1)^2=1$ and to $y=0$ with centre $C(x_C,y_C),y_C \leq 0$.
First of all, we consider this graph:
This is the same as proving that there exist no point $K$ such that $\overline{JK}=\overline{FK}$, with $x_J\geq x_L$. In particular, by Pythagora's theorem: $$\overline{FK}=\sqrt{(x_F-x_J)^2+(y_F-y_K)^2}$$ The lenght of the segment $JK$ is: $$\overline{JK}=-y_K$$ Now we want to show that $$\sqrt{(x_F-x_J)^2+(y_F-y_K)^2}>|y_K|$$ To do this, we square both sides, and we have: $$(x_F-x_J)^2+(y_F-y_K)^2>y_K^2\leftrightarrow (x_F-x_J)^2+y_F^2-2y_Fy_K>0$$ This is always true because, $y_K<0$ is $x_J\geq x_L$, so the term $2y_Fy_K$ is always positive as the other.
With this, we can conclude that th option A is incorrect.
Now, we study this graph:
In this case, the two points of tangents with $y=0$ and with $x^2+(y1)^2=1$ are the same $B(0,0)$, so every circunference with centre $C(0,y), y<0$ is correct.
This shows that the correct answer is B.
