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We have a Cartesian coordinate system with the points M (a,b) Q (4,2) and P (x,y) but I don't think you need P to solve this one, only M and Q. M is the middle of a circle with a radius r, and Q is a point on the circle (P is too, that's why I think P is redundant). What is the equation of this circle?

I thought it was: (4-a)^2 + (2-b)^2 = r^2 Is this correct my friends?

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    So is it allowed to keep asking seperate questions? Or will that be seen as laziness/spamming? I can assure you, I am not lazy, but I have a lot to learn.2012-09-13

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$M(a,b)$ is the center of the circle and $r$ is the radius.
And property of any circle is that every point on the circle are at the same distance $r$ from center.
So let $(x,y)$ be any point on the circle. So what is the distance of that point form the center ?? $d=\sqrt{(x-a)^2 + (y-b)^2} \quad \quad \text{Why ??}$ and we know that distance is same for every point on circle and is equal to $r$. So
$\sqrt{(x-a)^2 + (y-b)^2} = r$ $(x-a)^2 + (y-b)^2 = r^2$ Hence this is the equation of the circle with center at $(a,b)$ and radius $r$

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    @PhysicalEntity See if (x-a)^2 + (y-b)^2 > r^2 that means the distance of point $(x,y)$ from center is greater than $r$ so it must be outside the circle and similarly if that expression is less than $r^2$ then point is inside the circle.2012-09-14
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$(4-a)^2 + (2-b)^2 = r^2$, so you have to take the square root of the LHS:

$r = \sqrt ((4-a)^2 + (2-b)^2)$