For what value of c will the circle with equation x^2 + 8x + y^2 + 4y + c = 0 have a radius of length 7?
standard form of a circle = (x-a)^2 + (y-b)^2 = r^2 where (a,b) is the center and r is the radius
now, we have to account to complete the squares. since r = 7, r^2 = 49.
(x+4)^2 = x^2 + 8x + 16
(y+2)^2 + y^2 + 4y + 4
x^2 + 8x + 16 + y^2 + 4y + 4 + 49 = 0
this makes up a circle if you break it down into standard form... now lets add up what we have added onto it
49 + 4 + 16 = 49 + 20 = 69
+118609 Thanks guest,
I just want to have a quick look too.
For what value of c will the circle with equation x^2 + 8x + y^2 + 4y + c = 0 have a radius of length 7?
\( x^2 + 8x + y^2 + 4y + c = 0\\ x^2 + 8x + y^2 + 4y = -c\\ x^2 + 8x +16+ y^2 + 4y +4 = -c+16+4\\ (x+4)^2+(y+2)^2=20-c\\~\\ 20-c=49\\ 20-49=c\\ c=-29 \)
That is what I get 
