Question

How do you find the center and radius of the circle `x^2+y^2+4x-8y+4=0`?

Answer 1

centre = (-2 ,4) , radius = 4

Explanation:

The `color(blue)"general form of the 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 = `color(red)(|bar(ul(color(white)(a/a)color(black)(sqrt(g^2+f^2-c))color(white)(a/a)|)))`

`x^2+y^2+4x-8y+4=0" is in this form"`

To find the centre and radius, we require to identify g , f and c

By comparing the coefficients of 'like terms' in the given equation with the general form.

2g = 4 → g = 2 , 2f = -8 → f = -4 and c = 4

`rArr" centre" =(-g,-f)=(-2,4)`

and radius `=sqrt(2^2+(-4)^2-4)=sqrt(4+16-4)=4`

Alternatively Use the method of `color(blue)"completing the square"`

Add `(1/2"coefficient of x/y terms")^2" to both sides"`

`(x^2+4xcolor(red)"+4")+(y^2-8ycolor(red)"+16")=color(red)"4+16"-4`

`rArr(x+2)^2+(y-4)^2=16`

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.

By comparison of equation with standard form.

a = -2 , b = 4 and r = 4

Thus centre = (-2 ,4) and radius = 4