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

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Answer 1 · 2016-08-01T23:23:47

Centre is thus $(0, 6)$ $(0,6)$ and radius is $\sqrt{11}$ $sqrt(11)$

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.

$x^{2} + y^{2} - 12 y + 25 = 0$ $x^2+y^2-12y+25=0$

Subtract 25 from both sides:

$x^{2} + y^{2} - 12 y = - 25$ $x^2+y^2-12y = -25$

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

$x^{2} + y^{2} - 12 y + {(- 6)}^{2} = - 25 + {(- 6)}^{2}$ $x^2 + y^2 - 12y + (-6)^2 = -25 + (-6)^2$

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.

$x^{2} + {(y - 6)}^{2} = 11$ $x^2 + (y-6)^2 = 11$

This is now in the standard form

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

Centre is thus $(0, 6)$ $(0,6)$ and radius is $\sqrt{11}$ $sqrt(11)$

How do you find the center and radius for x2+y2−12y+25=0x^2 + y^2 -12y + 25= 0?