Find the center and radius of a circle whose equation is x^2+y^2+88=8y-22x??? Can someone please help me??? I have no idea how to do this!!

Question

aum · Answer

Group the "x" terms together and complete the square.
Group the "y" terms together and complete the square.
Express result in the form: (x-h)^2 + (y-k)^2 = r^2
Center will be: (h,k) and radius will be r.

anonymous · Answer

if the equation is of the from  x^2+y^2 +2gx +2fy +c =0
then centre of the circle is (-g,-f) and radius is sqrt(g^2+f^2 -c)
compare the given 
 x^2+y^2+88=8y-22x or ( x^2+y^2 +22x -8y +88=0) with the  x^2+y^2 +2gx +2fy +c =0
and identify f,g,c

anonymous · Answer

I still don't understand... What am I supposed to do to get  the answer?

anonymous · Answer

here u have 22 =2g means g =11
and 2f =-8     means f=-4   also c is 88  
so centre is (-11,4) and radius is  sqrt(  11^2 +4^2+88) =sqrt(225) = 15