How do you find an equation of the circle with center (2,4) and radius 3?

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Answer 1 · 2015-12-06T10:05:34

#x^2+y^2-4x-8y = -11#

If you know the coordinates of the center to be #(x_0,y_0)# and the radius to be #r#, then the equation of the circle with these center and radius is

#(x-x_0)^2 + (y-y_0)^2 = r^2#.

Plug your values and you have

#(x-2)^2 + (y-4) = 9#

You can either leave the equation as it is, or expand it:

#x^2-4x+4 + y^2-8y+16 = 9#

#x^2+y^2-4x-8y = -11#