How do you find the center and radius for #x^2+y^2+10x+6y-27=0#?

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Answer 1 · 2016-06-22T13:09:08

The center is #(-5,-3)# and radius is #sqrt61#.

If the equation is in the form

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

The center is #(h,k)# and radius is #r#.

Hence, let us convert the equation #x^2+y^2+10x+6y-27=0# to this form,

#x^2+10x+y^2+6y=27#

#hArr x^2+10x+25+y^2+6y+9=27+25+9#

#hArr (x+5)^2+(y+3)^2=61#

#hArr (x-(-5))^2+(y-(-3))^2=(sqrt61)^2#

Hence, the center is #(-5,-3)# and radius is #sqrt61#.