How do you find the center and radius of the circle given $x^2+y^2-6y-16=0$ ?

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Answer 1 · 2018-06-05T14:57:20

The process is complete the squares. See below

Remember that $(a+b)^2=a^2+2ab+b^2$ and
$(a-b)^2=x^2-2ab+b^2$

Then
$x^2+y^2-6x-16=(x-3)^2+y^2-16-9=(x-3)^2+y^2-25=0$

Then we have finally $(x-3)^2+y^2=25=5^2$

This is the equation of a circle centered in $(3,0)$ and radius 5

How do you find the center and radius of the circle given x^2+y^2-6y-16=0x^2+y^2-6y-16=0?