An analytic geometry question + algebra

Question

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?

Answer 1

$M(a,b)$ 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 $r$

Answer 2

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

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