A particular circle in the standard (x,y) coordinate plane has an equation of (x-5) squared + y squared = 38. What are the radius of the circle, in coordinate units, and the coordinates of the center of the circle?

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A particular circle in the standard (x,y) coordinate plane has an equation of (x-5) squared + y squared = 38. What are the radius of the circle, in coordinate units, and the coordinates of the center of the circle?

Mathematics
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The coordinates of the centre are (-1,0)
Could you give me an explanation of how you got your answer?
Also the + was actually supposed to be an = so excuse that my bad

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Other answers:

expand the equation given to get \[x ^{2}+y ^{2}-2x +63 = 0\]
general equation of a circle is x^2 + Y^2 +2gx + 2fy + c = 0
The equation is \[(x-5)^{2}+y ^{2}=38\]
Comparing the two we get g = -1 y=0 Also centre is equal to (-g.-f)
sorry the centre would be (1,0)
I think the center would be 5,0 because if we set x to =0 we get x=5 and since the y is only squared the y would automatically be 0. I just can't remember the radius. All i know is that it's the number the whole thing is equal to and i think it's square rooted.
Thanks though! I think i've got the answer :)

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