How do you find the equation of a circle in standard form given C(2,-4) and r=8?

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Answer 1 · 2016-12-19T06:50:20

Equation of a circle in standard form is $x^2+y^2-4x+8y-44=0$

If the center is $C(2,-4)$ and radius is $8$ ,

the point $(x,y)$ on the circle moves so that its distance from $C(2,-4)$ is always $8$

As distance of point $(x,y)$ from $C(2,-4)$ is $8$ , we have

$sqrt((x-2)^2+(y-(-4))^2)=8$

or $(x-2)^2+(y+4)^2=8^2$

or $x^2-4x+4+y^2+8y+16=64$

or $x^2+y^2-4x+8y-64+4+16=0$

or $x^2+y^2-4x+8y-44=0$
graph{x^2+y^2-4x+8y-44=0 [-19, 21, -13.68, 6.32]}