How to find the center and radius of #2x^2 + 2y^2 -8x +12y +8=0#?

1 Answer
Jan 29, 2016

centre = (2 , -3 ) and r = 3

Explanation:

The general equation of a circle is

# : x^2 + y^2 + 2gx + 2fy + c = 0#

For the given equation to be compared require to divide by 2 .

hence equation is # x^2 + y^2 - 4x + 6y + 4 = 0#

Comparing the equation to the general form.

then 2g = - 4 → g = - 2 and 2f = 6 →f = 3 , c=4

centre = ( - g , - f ) = (2 , - 3 )

and r = # sqrt(g^2 + f^2 - c ) = sqrt((-2)^2 + 3^2 - 4 ) = 3#