How do you find the center and radius for #y^2+ x^2-12y +10x+ 57=0#?

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Answer 1 · 2016-07-02T14:57:08

so radius 2: centre (-5,6)

complete the square -- TWICE!!

#y^2+ x^2-12y +10x+ 57=0#

#y^2 - 12y+ x^2 +10x+ 57=0#

#(y - 6)^2 - 36+ (x+5)^2 - 25+ 57=0#

#(y - 6)^2 + (x+5)^2 =36 +25 - 57#

#(y - 6)^2 + (x+5)^2 =2^2#

so radius 2: centre (-5,6)