How do you find the center and radius for $x^2+y^2-2x-8y+16=0$ ?

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Answer 1 · 2016-06-05T14:08:52

$C(1;4) and r=1$

Center coordinates are $(-a/2;-b/2)$ where a and b are the coefficients for x and y, respectively, in the equation;

$r=1/2sqrt(a^2+b^2-4c)$

where c is the constant term

so

$r=1/2sqrt(4+64-4*16)$
$r=1/2sqrt(4)$
$r=1/2*2=1$

How do you find the center and radius for x^2+y^2-2x-8y+16=0x^2+y^2-2x-8y+16=0?