How do you find the center and radius for #x^2 + y^2 -12y + 25= 0#?

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Answer 1 · 2016-08-01T23:23:47

Centre is thus #(0,6)# and radius is #sqrt(11)#

We want to complete the squares. As there is no x term of degree one we leave it as is and focus on the y.

#x^2+y^2-12y+25=0#

Subtract 25 from both sides:

#x^2+y^2-12y = -25#

Square half the coefficient of #y# and add to both sides:

#x^2 + y^2 - 12y + (-6)^2 = -25 + (-6)^2#

Recognise that because the constant term is half the coefficient we can convert the left hand side to squared form and simplify the right hand side.

#x^2 + (y-6)^2 = 11#

This is now in the standard form

#(x-a)^2 + (y-b)^2 = r^2# for circle centre #(a,b)# and radius #r#.

Centre is thus #(0,6)# and radius is #sqrt(11)#