What is the center of #(x+1)^2 + (y-3)^2=9#?

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Answer 1 · 2016-10-31T17:45:35

#(-1, 3)#

Rewrite the equation as:

#(x - -1)^2 + (y - 3)^2 = 9#

as the center is found by observation #(-1,3)#

Answer 2 · 2016-10-31T18:05:38

The center of the circle represented by given equation is #(-1,3)# and its radius is #3#.

An equation of the form #(x-h)^2+(y-k)^2=r^2#, is the equation of a circle of radius #r#, with a center at #(h,k)#.

Hence as the equation

#(x+1)^2+(y-3)^2=9#

#hArr (x-(-1))^2+(y-3)^2=9=3^2#

and hence the center of the circle represented by this equation is #(-1,3)# and radius is #3#.