How do you find the center and radius of the circle given #3x^2+3y^2+12x-6y+9=0#?

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Answer 1 · 2016-11-23T16:39:43

The center and radius are related to the coefficient of the equation.

First divide all the equation by three to put in simple form:

#x^2 + y^2 + 4x -2y + 3 =0#

Now consider the general equation of the circle is:

#(x - x_o)^2 + (y- y_o)^2 - R^2 = 0#

where the center is the point #(x_o, y_o)# and #R# is the radius.

Expanding:

#x^2 - 2x x_o +x_o^2 + y^2 - 2yy_o +y_o^2 - R^2 = 0#

Comparing similar terms, you can easily see that:

#-2x_o = 4 =>x_o = -2#
#-2y_o = -2 => y_o = 1#
#x_o^2 +y_o^2 -R^2 = 3 => R = sqrt(2)#

graph{x^2+y^2+4x-2y+3=0 [-10.375, 9.625, -4.08, 5.92]}