Which of the following are partitions of $\mathbb R^2$ Is my answer correct?
Can someone provide me better explanations for (a) ,(c) and (d)?

Which of the following collections of subsets of the plane $\Bbb R\times\Bbb R$ are partitions?
$(a)$ $\left\{\left\{(x,y)\,:\,x+y=c\right\}\,:\,c\in\Bbb R\right\}$
$(b)$ The set of all circles in $\Bbb R\times\Bbb R$
$(c)$ The set of alla circles in $\Bbb R\times\Bbb R$ centered at $(0,0)$, together with $\{(0,0)\}$
$(d)$ $\left\{\left\{(x,y)\right\}\,:\,(x,y)\in\Bbb R\times\Bbb R\right\}$

a) will be a partition as we can cover $\mathbb R^2$ with straight lines.
b) will not be a partition as elements of this set are not disjoint.
c) will be a partition as we can cover $\mathbb R^2$ with circles having origin as center.
d) will be a partition as they are equivalence class of relation $(x,y) R (x',y')$ if $(x,y) = (x',y')$, equivalence classes will be singletons only
 A: First of all, your answers are correct.
a) Obviously these sets cover $\mathbb{R}^2$. Assume we have two sets which are not disjoint, i.e. $(x,y) \in A_{c_1} \cap A_{c_2}$. Then we have $c_1 = x + y = c_2$. Thus $A_{c_1} = A_{c_2}$ and therefore it really is a partition.
b) You should include an example (which is very easy).
c) Again you have to show that two circles with origin $(0,0)$ are either disjoint or equal.
A: Your answers are correct.
Sometimes, this criterion turns out to be pretty handy:

Theorem: Let $Q$ be a set and let $S\subseteq \mathcal P(Q)$. The following are equivalent:
  
  
*
  
*$S$ is a partition of $Q$
  
*There exist a set $R\neq\emptyset$ and a surjective function $f:Q\to R$ such that $S=\left\{f^{-1}\{c\}\,:\,c\in R\right\}$.
  
  
  Where $f^{-1}\{c\}:=\{ q\in Q\,:\, f(q)=c\}$. 

This provides a straight-forward argument to prove $(a)$, since $$+:\Bbb R\times\Bbb R\to \Bbb R$$ is surjective: you need only use $Q=\Bbb R\times\Bbb R$, $S=\text{your candidate}$, $f=(\bullet)+(\bullet)$ and $R=\Bbb R$
It can also be used for $(c)$, with the function $$\lVert\bullet\rVert:\Bbb R\times\Bbb R\to \Bbb R$$
And with $(d)$ as well, using $id_{\Bbb R\times\Bbb R}$.
