Question 1089921

{{{x^2 +y^2 +4x+4y-1=0}}}

to find the radius of the circle, write your equation in form:

{{{(x-h)^2+(y-k)^2=r^2}}} where {{{h}}} and {{{k}}} are coordinates of the center, and {{{r}}} is the radius

{{{(x^2 +4x)+(y^2+4y)-1=0}}}...complete squares

{{{(x^2 +4x+b^2)-b^2+(y^2+4y+b^2)-b^2-1=0}}}...for {{{x}}} part {{{a=1}}} and {{{2ab=4}}}=>{{{2*1b=4}}}=>{{{b=2}}}
for {{{y}}} part {{{a=1}}} and {{{2ab=4}}}=>{{{2*1b=4}}}=>{{{b=2}}}

{{{(x^2 +4x+2^2)-2^2+(y^2+4y+2^2)-2^2-1=0}}}

{{{(x+2)^2-4+(y+2)^2-4-1=0}}}

{{{(x+2)^2+(y+2)^2-9=0}}}

{{{(x+2)^2+(y+2)^2=9}}}

{{{(x+2)^2+(y+2)^2=3^2}}}=>the radius of the circle is {{{r=3}}}