How do you find the center and radius of the the circle #x^2 + y^2 + 6/5x - 8/5y - 8 = 0#?

1 Answer
Feb 13, 2016

Answer:

First factorise the original equation, so that you have it in the form of a circle equation:
#(x+3/5)^2+(y-4/5)^2=9#
From this we can see that the radius #=sqrt(9)=3#, the centre x-coordinate is -3/5=-0.6 and the centre y-coordinate is 4/5=0.8.

Explanation:

Starting with the original equation:
#x^2 + y^2 + 6/5 x - 8/5 y - 8 = 0#
which equals:
#x^2 + 6/5 x + y^2 - 8/5 y + (1 - 9)#

= #x^2 + 6/5 x + 9 /25 +y^2 - 8/5 y +16/25 - 9#

Now factorise the above equation, so that you have it in the form of a circle equation:
#(x+3/5)^2+(y-4/5)^2=9#

The circle equation is:
#(x-x_0)^2+(y-y_0)^2=r^2#

Where r is the radius. The centre x point and the centre y point are #x_0 and y_0# respectively.

From this we can see that the radius #=sqrt(9)=3#, the centre x-coordinate is -3/5=-0.6 and the centre y-coordinate is 4/5=0.8.