How do you find the center and radius for $x^2 - 8x + y^2 - 4y – 5 = 0$ ?

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Answer 1 · 2016-06-10T09:22:21

Center is $(4,2)$ and radius is $5$

In the equation of a circle in form

$x^2+y^2+2gx+2fy+c=0$

The center of circle is at $(-g,-f)$ and radius is $sqrt(g^2+f^2-c)$

In equation $x^2-8x+y^2-4y-5=0$ , $g=-4$ , $f=-2$ and $c=-5$

Hence, center is $(4,2)$ and radius is

$sqrt(4^2+2^2-(-5))=sqrt25=5$

graph{x^2-8x+y^2-4y-5=0 [-7.08, 15.42, -3.345, 7.905]}

How do you find the center and radius for x^2 - 8x + y^2 - 4y – 5 = 0x^2 - 8x + y^2 - 4y – 5 = 0?