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

1 Answer
Shwetank Mauria
Dec 19, 2016

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

Explanation:

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]}

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