What is the center and radius of the circle with equation #(x-5)^2+(y+3)^2=16#?

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Answer 1 · 2016-06-14T15:59:33

#(x-5)^2+(x+3)^2=16#

The center of the circle is #(5,-3)# the inverse of both #h# and #k#.

The radius is the #sqrt16# which is #r=4#.

The equation of the circle is

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

Where the center of the circle is (-h,-k)

for the equation

#(x-5)^2+(x+3)^2=16#

Therefore, the center of the circle is #(5,-3)# the inverse of both #h# and #k#.

The radius is the #sqrt16# which is #r=4#.