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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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As a rule, we don't like questions where you are constantly adding to the question after an answer has been given. This is not a tutoring site, it is a question/answer site. – Thomas Andrews Sep 13 '12 at 19:48
How can one learn from just one question.. – PhysicalEntity Sep 13 '12 at 19:51
If you have more than one question, you post more than one question. You don't keep adding to the original question. – Thomas Andrews Sep 13 '12 at 19:55
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. – PhysicalEntity Sep 13 '12 at 19:56
up vote 2 down vote accepted

$M(a,b)$ 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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Thank you, so if for instance, the center point becomes (-5,2) and the radius 13 it is just filling in a,b and r in the formula? – PhysicalEntity Sep 13 '12 at 20:15
@PhysicalEntity yes, exactly – Saurabh Sep 13 '12 at 20:17
Last question, I promiss: I have filled in (2,-9) in the formula I got and I got 141 as answer instead of 140. How would I know if this point (2,-9) is in the circle or out of the circle? If it is bigger than the answer should be, is it in or out? Drawing it I figured it is in the circle, but what if I weren't allowed to draw? – PhysicalEntity Sep 13 '12 at 20:38
@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. – Saurabh Sep 14 '12 at 2:47

$(4-a)^2 + (2-b)^2 = r^2$, so you have to take the square root of the LHS:

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

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