Find the center and radius of a circle whose equation is #(x+5)^2+(y-3)^2=81#?

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Answer 1 · 2017-01-08T15:07:56

Answer is #(2)# i.e. center is #(-5,3)# and radius is #9#.

The equation of a circle whose center is #(h,k)# and radius is #r# is

#(x-h)^2+(y-k)^2=r^2#............(A)

As in #(x+5)^2+(y-3)^2=81#, we have plus sign in first term,

we change #(x+5)# to #(x-(-5))#

Hence #(x+5)^2+(y-3)^2=81# is equivalent to

#(x-(-5))^2+(y-3)^2=9^2#

Comparing this with (A), we have #h=-5#, #k=3# and #r=9#

Hence center is #(-5,3)# and radius is #9# and

Answer is #(2)# i.e. center is #(-5,3)# and radius is #9#.