Question :

Question

find (h,k) and radius
x^2 + y^2 +1/2x + 2y +1/16 = 0

According to the equation :
(x−h)^2+(y−k)^2 = r^2

x^2 + y^2 +1/2x + 2y +1/16 = 0
(x^2+1/2x+1/16)+(y^2+2y+1)=−1/16+1/16+1
(x+1/4)^2+(y+1)^2=1

so (h,k) = (-1/4, -1)
and r = 1

theres no answer to even number from the textbook, i would like to clarify if im doing it correctly. Thanks a lot :D

Answer 1

looks good

Answer 2

thanks