How do you find the center and radius for $x^2 + y^2 - 8x - 2y - 8 = 0$ ?

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Answer 1 · 2016-06-01T18:19:31

The center is $(4, 1)$ and the radius $3$ .

$0 = x^2+y^2-8x-2y-8$

$= x^2-8x+16 + y^2-2y+1 - 9$

$= (x-4)^2+(y-1)^2-3^2$

Add $3^2$ to both ends and transpose to get:

$(x-4)^2+(y-1)^2 = 3^2$

This is in the form:

$(x-h)^2+(y-k^2) = r^2$

which is the general form of the equation of a circle with centre $(h, k) = (4, 1)$ and radius $r=3$

How do you find the center and radius for x^2 + y^2 - 8x - 2y - 8 = 0x^2 + y^2 - 8x - 2y - 8 = 0?