Question

What is the center and radius of the circle with equation `36x^2 - 36x + 36y^2 - 24y = 131`?

Answer 1

The General equation of circle is : `x^2 + y^2 + 2gx + 2fy + c = 0`

Explanation:

So,
Dividing all terms by 36 ( To remove the coefficient of x^2 and y^2)
We get,
`x^2 - x + y^2 + 2/3 y = 131/36`
Now
`x^2 - x + y^2 + 2/3 y- 131/36 = 0`

Comparing the given equation with the general equation of circle we get,

2g = `-1`
or g = `-1/2`

Again,
2f = `2/3`
or f = `1/3`

And,

c= `-131/36`

Thus,
Center of the circle (h,k) = (`-g ,-f`)
= (`1/2 , -1/3`)
And
Radius = `sqrt(g^2+f^2-c)`