AmTran_Bus
  • AmTran_Bus
How to find the center of a circle (x-h)^2 + (y-k)^2 =r^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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AmTran_Bus
  • AmTran_Bus
Is it always the opposite of whatever the h and k values are?
dape
  • dape
Yes it is. Do you know why?
AmTran_Bus
  • AmTran_Bus
I would love for you to tell me.

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AmTran_Bus
  • AmTran_Bus
So, for example, the center of (x+2)^2 + (y+3)^2 ia -2,-3?
anonymous
  • anonymous
yup
anonymous
  • anonymous
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
  • AmTran_Bus
Thanks.
dape
  • dape
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
  • AmTran_Bus
Thanks Dape!

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