Cardinality of a set of circles Find the cardinality of a set of circles in the plane whose centre has rational coordinates and whose radius is the square root of a prime number.
 A: Let $A$ be the set of such circles, for every $x\in A$ let $c_x=(h_x,k_x)$ denote the center of $x$, and $r_x$ denote the radius of $x$. Now define $f:X\rightarrow \mathbb{Q}\times \mathbb{Q}\times \mathbb{P}$ as $x\mapsto (c_x,r_x)=(h_x,k_x,r_x)$. This is clearly a bijection. Now, assuming Axiom of Choice, since $$\mid \mathbb{Q}\mid=\mid \mathbb{P}\mid $$, then $\mid X \mid =\mid \mathbb{Q}\times \mathbb{Q}\times \mathbb{P} \mid = \aleph_0$
Note: Here $\mathbb{P}$ denotes the set of prime numbers
A: They are trying to confuse you with complex seeming conditions.
Thing is, (if we let $\mathbb P$ be the notation for the set of primes) then
Each circle can be described by $(a,b,p)\in \mathbb Q \times\mathbb Q\times \mathbb P \leftrightarrow C:= \{(x,y)| (x-a)^2 + (y-b)^2 = p\}$
If we want to be technically and jargon heavy that is a bijection between $\mathbb Q \times \mathbb Q\times \mathbb P $ with the set of all your circles.
So the cardinality you seek, is simply the cardinality of $\mathbb Q\times \mathbb Q\times \mathbb P$.
And you should know the cardinality of $\mathbb Q$ and you should probably be able to state what the cardinality of $\mathbb P$ (which is an infinite subset of $\mathbb N$) is, and there is proposition you should know about cardinalities of cross products of sets  you  know the cardinality of....
