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

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Answer 1 · 2018-06-05T14:57:20

The process is complete the squares. See below

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

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

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

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

How do you find the center and radius of the circle given x2+y2−6y−16=0x^2+y^2-6y-16=0?