How to find the center of a circle

(x-h)^2 + (y-k)^2 =r^2

Question

How to find the center of a circle

(x-h)^2 + (y-k)^2 =r^2

amtran_bus · Answer

Is it always the opposite of whatever the h and k values are?

dape · Answer

Yes it is. Do you know why?

amtran_bus · Answer

I would love for you to tell me.

amtran_bus · Answer

So, for example, the center of (x+2)^2  + (y+3)^2 
ia -2,-3?

anonymous · Answer

yup

anonymous · Answer

basically re-write the equation of that circle as (x-(-2))^2 + (y-(-3))^2
and then you see that the centre is infact (-2, -3)..

amtran_bus · Answer

Thanks.

dape · Answer

Think of a curve, any curve. We can write the curve you are thinking about as the graph of an equation like this:
$$f(x,y)=0$$
Don't be worried, just think of this being any stuff involving x's and y's on the left. And if you put in a x and y so that the left hand side becomes zero, it is a point on the curve. As an example, let's say the point (3,5) is on the graph, in our equation this just means $f(3,5)=0$. Now we think about what happens when we do $f(x-1,y)=0$. Now the point (4,5) is a point on the graph because $f(4-1,5)=f(3,5)=0$.

So you see that the point (3,5) moved a step to the RIGHT to (4,5) when we shifted x "down" by 1 in the equation. So if you do $f(x+u,y+u)=0$, the curve will be shifted exactly $u$ "steps" to the left and $v$ "steps" downwards. So the curve moves opposite, in a way, to what we do to x and y.

For the circle you can just put $f(x,y)=x^2+y^2-r^2=0$. In just the same way as above, we can get your original equation by doing $f(x-h,y-k)=0$, so we have a circle shifted $h$ steps to the RIGHT and $k$ steps UP, in a way "opposite" of the h and k values.

amtran_bus · Answer

Thanks Dape!