Solving $\sqrt{(x-2)^2 + (y-1)^2} = \sqrt 2$ What is the answer of this:
$\sqrt{(x-2)^2 + (y-1)^2} = \sqrt 2$
 A: There is not a single solution, but rather several possible $(x,y)$ that satisfy that relation. If you don't recognize it right off - let's see what we can think of.
$\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$ is the distance formula between points $(x_1, y_1)$ and $(x_2, y_2)$. So you are asking for the set of $(x,y)$ that are distance $\sqrt 2$ away from the point $(2,1)$.
The locus of points equidistant to a single point is a circle. So the set of solutions form a circle in the plane.
A: That's just an equation, not a question. But if there are some written instructions attached to the equation, such as

Identify the set of solutions to
  $ \sqrt{(x-2)^2 + (y-1)^2} = \sqrt 2 $

then almost certainly the expected answer is not just another equation, but a description of the solution set in words, such as

The solutions are the points on the circle of radius ___ centered at the point ( ___ , ___ ).

A: $\textbf{Hint}$ : First, try to relate this to the pythagorean theorem. Also note.
$\sqrt{(x - 2)^2 + (y - 1)^2} = \sqrt{2}$
$(x - 2)^2 + (y - 1)^2 = 2$
Think about circles. 
