How do you rewrite the following equation in the center-radius form of the equation of a circle #x^2 - 4x + y^2 - 4y + 4 = 0#?

1 Answer
Jun 26, 2016

Answer:

#(x-2)^2+(y-2)^2=4#

Explanation:

The 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)|)))#

centre = (-g ,-f) and radius (r)=#sqrt(g^2+f^2-c)#

If we compare like terms in the given equation to the general form.

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

hence centre = (2 , 2) and r #=sqrt((-2)^2+(-2)^2-4)=2#

The equation of the circle in #color(blue)"centre-radius form"# 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.

substitute a = 2 , b =2 and r =2 into this equation

#rArr(x-2)^2+(y-2)^2=4" is the equation"#