Why is the centre of a circle: $(x + a)^2 + (y + b)^2 = c^2$?

Question

1. Why is the centre of the circle:

   $$(x + a)^2 + (y + b)^2 = c^2,$$

$(-a,-b)$?

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2. What is a good way of remembering that the centre of the circle:

   $$(x + a)^2 + (y + b)^2 = c^2$$

is $(-a,-b)$ and not $(a,b)$?

Answer 1

The equation relating points lying on the circle centre $(a,b)$ and radius $c$ is given by $$\sqrt{(x-a)^2+(y-b)^2}=c$$ where the left hand side gives the distance of the point $(x,y)$ from the centre and the right hand side gives the radius. For the distance from point $(x,y)$ the centre to be equal to $0$ we need $(x,y)=(a,b)$ hence the coordinates of the centre must be $(a,b)$. The equation above can be rewritten as $$(x-a)^2+(y-b)^2=c^2$$ $$(x+(-a))^2+(y+(-b))^2=c^2$$ which gives the same form you mention.

Answer 2

The euclidean distance from the point $M(x,y)$ to the point $C(a,b)$ is the length of the vector $$\overrightarrow{CM}=\overrightarrow{OM}-\overrightarrow{OC}=x\,\vec i+y\,\vec j-(a\,\vec i+b\,\vec j)=(x-a)\,\vec i+(y-b)\,\vec j;$$ whence the formula for the length in an orthonormal basis.

Answer 3

I would instead say that the centre of circle $$(x-a)^2+(y-b)^2=c^2$$ is $(a,b)$.

Notice how this substitution requires $c^2=0\implies c=0$.

Answer 4

It might be useful to remember that in just one dimension, the distance between two points on the number line is (the absolute value of) their difference, not their sum. For instance, if you have a point at $a = -2$, and another point at $b = 4$, then the distance between them is $|b - a| = |4 - (-2)| = 6$, not $|4 + (-2)| = 2$.

In fact, we could express the distance $c$ between them using the equation $c = \sqrt{(a-b)^2}$, and it would be perfectly valid. We only normally don't do that because in one dimension, it isn't necessary. But it would make the generalization to $n$ dimensions clearer.