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

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