# Logic | consistency model question

We define the circle with the center point – $$(a,b)$$ and his radius – $$r>0$$ as:
$$B(a,b,r)={langle x,y rangle in mathbb{R}^2:(x-a)^2+(y-b)^2le r^2}$$
Let $$A$$ be the set of the plane’s circles:
$$A={B(a,b,r):a,bin mathbb{R},rin(0,infty )}$$
Let $$mathcal{L}$$ be the vocabulary which contains a binary relation $$S$$, and in addition contains for each circle – $$B(a,b,r)in A$$ constant symbol $$bar c_{a,b,r}$$. Let $$M$$ a model to interpret $$mathcal{L}$$, defined by:$$M=langle A,Srangle$$ When $$S$$ defined to be the $$subseteq$$ relation between the circles ($$S(x,y)$$ meaning that $$xsubseteq$$y). The model also interpret $$bar c_{a,b,r}$$ as a circle $$B(a,b,r)$$. Let $$T=Th(M)$$ the theory of $$M$$. We define $$T_1$$, $$T_2$$ in the vocabulary $$mathcal{L}^+=mathcal{L}cup{d}$$, where $$d$$ is a new constant symbol:
$$T_1=Tcup{forall x (S(x,d))}$$
$$T_2=Tcup{S(d,c_{a,b,r}):a,bin mathbb{R},rin(0,infty )}$$
Now, for $$T_1, T_2$$ I need to check if:

• The theory is inconsistent
• The theory is is consistent, and it has a model which is an expansion of $$M$$ to $$mathcal{L}^+$$.
• The theory is consistent, but it does not have a model which is an expansion of $$M$$ to $$mathcal{L}^+$$.

My attempt: I wanted to think about the meaning before diving into giving $$d$$ interpretation and so on. Therefore, for $$T_1$$, it means that for every circle $$x$$ is within the biggest circle. Now in the world of $$B(a,b,r)$$ there isn’t the biggest one, so if I am appealing to the compatible sentence, I can choose a circle that will contain all circles within him. Therefore $$T_1$$ is consistent but she hasn’t an expansion of $$M$$ to $$mathcal{L}^+$$.
For $$T_2$$: the additional sentence means that there is the smallest circle, which impossible, also if we applying the compatible theorem, because finite groups of circles, can scatter around the cartesian coordinate system. Hence, we have that $$T_2$$ is inconsistent.

I am not sure that my way is right. Thus, I will be glad to get some help from you. Thank you!