Where is the center of the circle given by the equation #(x - 3)^2 + (y + 2)^2 = 9#?

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Answer 1 · 2018-05-25T07:45:17

the centre is#" "(3,-2)#

an eqn of a circle is given by

#(x-a)^2+(y-b)^2=r^2--(1)#

where the centre is #(a,b)#

and the radius is #r#

for

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

comparing with #(1)#

the centre is#(3,-2)#

as an extra we see the radius is

#r=sqrt9=3#