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

Question

3231 views around the world

You can reuse this answer
Creative Commons License

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 $sqrt(11)$

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