How do you find the center and radius of the circle given 3x^2+3y^2+12x-6y+9=03x2+3y2+12x6y+9=0?

1 Answer
Nov 23, 2016

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

Explanation:

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

x^2 + y^2 + 4x -2y + 3 =0x2+y2+4x2y+3=0

Now consider the general equation of the circle is:

(x - x_o)^2 + (y- y_o)^2 - R^2 = 0(xxo)2+(yyo)2R2=0

where the center is the point (x_o, y_o)(xo,yo) and RR is the radius.

Expanding:

x^2 - 2x x_o +x_o^2 + y^2 - 2yy_o +y_o^2 - R^2 = 0x22xxo+x2o+y22yyo+y2oR2=0

Comparing similar terms, you can easily see that:

-2x_o = 4 =>x_o = -22xo=4xo=2
-2y_o = -2 => y_o = 12yo=2yo=1
x_o^2 +y_o^2 -R^2 = 3 => R = sqrt(2)x2o+y2oR2=3R=2

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