Question

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

Answer 1

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.