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What is the answer of this:

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

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    @PhysicalEntity that's right.2012-09-13

3 Answers 3

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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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    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...2012-09-13
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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.

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

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

Think about circles.

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    Thank you, you confirmed what I thought it. I am sometimes a little hesitant to ask these kind of questions on a forum with all these people who are extremely good at math.. but I hope to improve my level of math by frequently asking questions and I hope I don't bother you guys to much..2012-09-13