What is the center of the circle with equation #(x + 1)^2 + (y + 3)^2 = 9/4#?

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Answer 1 · 2016-04-02T01:14:57

The center is at #(-1, -3)#

From the given equation #(x+1)^2+(y+3)^2=9/4#

this is comparable already to the Center-Radius form of the equation of the circle

#(x-h)^2+(y-k)^2=r^2# and we can adjust the given as follows

#(x+1)^2+(y+3)^2=9/4#

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

and clearly, the center is at #(-1, -3)#

God bless.....I hope the explanation is useful.