# Solving $\sqrt{(x-2)^2 + (y-1)^2} = \sqrt 2$

What is the answer of this:

$\sqrt{(x-2)^2 + (y-1)^2} = \sqrt 2$

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Not sure what you are asking "what's the answer". You have an implicit solution... so that's the answer. – Euler....IS_ALIVE Sep 13 '12 at 19:23
Well I would say you first have to square the answers on both sides, but I'm not sure, that's why I'm asking you guys. – PhysicalEntity Sep 13 '12 at 19:24
make the square in both sides.$a^2=b^2$ – HipsterMathematician Sep 13 '12 at 19:24
@PhysicalEntity that's right. – HipsterMathematician Sep 13 '12 at 19:25

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.

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Thank you, you're right, my final answer is x^2 - 4x + y^2 - 2y = -3 .. is this correct? It looks correct when I plot it on wolfram alpha – PhysicalEntity Sep 13 '12 at 19:31
@PhysicalEntity: I don't think that's a very good "final answer", since it's much clearer from your original equation that the solution set is a circle... – Hans Lundmark Sep 13 '12 at 19:34
@ Hans Yes you are right, but I that was our assignment.. I know, high school maths for you guys is probably extremely annoying and boring (and incorrect). – PhysicalEntity Sep 13 '12 at 19:35
It depends, what is considered an answer. I'd prefer to give the set of solutions e.g. as $\{(2+\sqrt 2 \cos t, -1+\sqrt 2 \sin t)\mid t\in\mathbb R\}$ instead of meerely transforming the given algebraic to a polynomial equation... – Hagen von Eitzen Sep 13 '12 at 19:45

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 ( ___ , ___ ).

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$\textbf{Hint}$ : First, try to relate this to the pythagorean theorem. Also note.

$\sqrt{(x - 2)^2 + (y - 1)^2} = \sqrt{2}$

$(x - 2)^2 + (y - 1)^2 = 2$