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

##### 1 Answer

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#