How do you find the equation of the circle given Center (8,-8), Radius 1/3?

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Answer 1 · 2016-09-17T10:43:26

#9x^2+9y^2-144x+144y+1151=0#

Circle with the given center at #(8,-8)# and radius #1/3#,

is the locus of a point that moves so that it is always at a distance of #1/3# from point #(8,-8)#

Hence equation is #(x-8)^2+(y-(-8))^2=(1/3)^2#

or #x^2-16x+64+y^2+16y+64=1/9#

or #9x^2+9y^2-144x+144y+9xx(64+64)-1=0#

or #9x^2+9y^2-144x+144y+1151=0#

graph{9x^2+9y^2-144x+144y+1151=0 [4.894, 9.894, -8.68, -6.18]}