Question

How do you find the center and radius of the circle `x^2 + y^2 + 4x - 4y - 1 = 0`?

Answer 1

centre = (-2 ,2) , radius = 3

Explanation:

The `color(blue)"general equation of a circle"` is

`color(red)(|bar(ul(color(white)(a/a)color(black)(x^2+y^2+2gx+2fy+c=0)color(white)(a/a)|)))`

with centre = `color(red)(|bar(ul(color(white)(a/a)color(black)(((-g,-f)))color(white)(a/a)|)))`

and radius (r) = `color(red)(|bar(ul(color(white)(a/a)color(black)(sqrt(g^2+f^2-c))color(white)(a/a)|)))`

the equation `x^2+y^2+4x-4y-1=0" is in this form"`

and by comparing the coefficients of like terms we get the values for g ,f and c.

`2g=4rArrg=2,2f=-4rArrf=-2" and " c=-1`

`rArr" centre"=(-g,-f)=(-2,2)`

and r = `sqrt(g^2+f^2-c)=sqrt(2^2+(-2)^2-(-1))=sqrt9=3`
`color(blue)"---------------------------------------------------------------------------"`

Alternatively the method of `color(blue)"completing the square"` can be used to obtain the equation.

`(x^2+4xcolor(red)(+4))+(y^2-4ycolor(red)(+4))-1=color(red)(4)+color(red)(4)`

`rArr(x+2)^2+(y-2)^2=9`

The`color(blue)" standard form of the equation of a circle"` is

`color(red)(|bar(ul(color(white)(a/a)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(a/a)|)))`
where (a ,b) are the coordinates of the centre and r, the radius.

`rArr"centre"=(-2,2)" and " r=3`