How many points are possible in the plane that are the same distance from all three points. 

P, Q and R are the points on a plane 

  
  
*
  
*How many points are possible in the plane that are equidistant from both P and Q?
  
*If R is equidistant from both P and Q, then how many points are possible in the plane that are the same distance from all three
  points. 
  
*If R does lie on line PQ but not equidistant from P and Q, then how many  points are possible in the plane that are the same distance from
  all three points.
  
*If R does not lie on line PQ, then how many points are possible in the plane that are the same distance from all three points.
  

For $1$, I find infinite points in up and down through the middle point of line PQ.
For $2$, I find one point
For $3$, I find no point
For $4$, I find only one point
Though this all is my guess after sketching lines on paper. Can anyone explain how to solve these?
 A: For the first problem, the set of points equidistant from $P$ and $Q$ is the perpendicular bisector of the line $PQ$. There are infinitely many points on this line, which thus satisfy the desired property.
For the last problem, $\triangle PQR$ is a non-degenerate triangle. Hence, there is exactly one point equidistant from the three given points, the circumcentre of $\triangle PQR$.
The second and third problems are exceptions to the answer for problem 4. If the three points lie on a line – and $R$ doesn't have to be in the middle of the line $PQ$ – the triangle degenerates into a line and no point on the plane will be equidistant from all three points. (Exception to the exception: if $R=P$ or $R=Q$ then the midpoint of $PQ$ is equidistant from all three points.)
A: I assume that the points $P$, $Q$, $R$ are all distinct. It's easy to enumerate the degenerate cases where two or more of them coincide.
Your answer to the first question is correct: the set of points equidistant from $P$ and $Q$ is the perpendicular bisector of the segment $\overline{PQ}$, namely the line perpendicular to $\overline{PQ}$ and crossing it at its midpoint. 
Consequently, if a point is supposed to be equidistant from $P$, $Q$ and $R$, it must belong to three lines simultaneously: the perpendicular bisector of $\overline{PQ}$, that of $\overline{PR}$ and that of $\overline{QR}$.
Now, if $PQR$ is an actual triangle, then the perpendicular bisectors of the sides indeed meet at a single point: the circumcenter. And of course, the circumcenter is indeed the unique point equidistant from the three vertices of the triangle.
If $PQR$ is not a triangle, meaning if $P,Q,R$ lie on a single line, then the perpendicular bisectors are parallel and there's no equidistant point. 
Therefore, the condition for there being an equidistant point from three distinct point $P,Q,R$ is that the points don't lie on a single line. If they don't, there's a unique such point. If they do, there's no such point. 
This, then, has nothing to do with whether or not $R$ is itself equidistant from $P$ and $Q$. It does depend on whether $R$ lies on the line $PQ$ or not.
