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?

Question

anonymous · Answer

The coordinates of the centre are (-1,0)

anonymous · Answer

Could you give me an explanation of how you got your answer?

anonymous · Answer

Also the + was actually supposed to be an = so excuse that my bad

anonymous · Answer

expand the equation given to get
 $$x ^{2}+y ^{2}-2x +63 = 0$$

anonymous · Answer

general equation of a circle is 
x^2 + Y^2 +2gx + 2fy + c = 0

anonymous · Answer

The equation is $$(x-5)^{2}+y ^{2}=38$$

anonymous · Answer

Comparing the two we get g = -1 y=0
 Also centre is equal to (-g.-f)

anonymous · Answer

sorry the centre would be (1,0)

anonymous · Answer

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.

anonymous · Answer

Thanks though! I think i've got the answer :)