Which is the center of the circle whose equation is (x - 3)2 + (y - 4)2 = 121?

Question

geometry_hater · Answer

I dont know how to find centers of circles

amistre64 · Answer

(x-Cx)^2 + (y-Cy)^2 = r^2

amistre64 · Answer

its not so much how to find it, but how to retrieve them from the equation

amistre64 · Answer

(Cx,Cy) is the center

geometry_hater · Answer

Ok i still dont know what to do for the problem

amistre64 · Answer

me either ....

agreene · Answer

$$ (x - 3)^2 + (y - 4)^2 = 121$$
if that is your equation then the center is at:
(3,4)
because it is in general form, like amistre64 stated:
$$(x-Cx)^2+(y-Cy)^2=r^2$$
where Cx and Cy are your central coordinates (note the negatives in the general form, meaning if it were (x+3)^2+(y+4)^2 it would be centered at (-3,-4))

geometry_hater · Answer

so -3 = the x's and -4 equals the y's???

agreene · Answer

No, it is asking for the center of the circle. If the equation is the same as the one I posted then the center is
(3,4)
because the equation is in general form, and you can just look at the equation and get that information.