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

looks good

thanks

To find the coordinates of the center of a circle (h, k) and its radius r from the given equation, you need to rewrite the equation in the standard form of a circle equation as (x−h)^2+(y−k)^2 = r^2.

Now let's go through the solution step by step:

1) Start with the given equation:
x^2 + y^2 +1/2x + 2y +1/16 = 0

2) Rearrange the equation to group the x and y terms together:
x^2 + 1/2x + y^2 + 2y + 1/16 = 0

3) Complete the square for x terms:
x^2 + 1/2x + (1/2*1/2) - (1/2*1/2) + y^2 + 2y + 1/16 = -1/16

4) Complete the square for y terms:
x^2 + 1/2x + 1/4 - 1/4 + y^2 + 2y + (2/2)^2 - (2/2)^2 + 1/16 = -1/16

5) Simplify the equation:
(x + 1/4)^2 + (y + 1)^2 + 1/16 - 1/4 - 1/4 - 1/16 = -1/16

6) Combine like terms on the right side:
(x + 1/4)^2 + (y + 1)^2 = 1

Now we have the equation in the standard form as (x−h)^2+(y−k)^2 = r^2, where h = -1/4, k = -1, and r^2 = 1.

So the center of the circle is (-1/4, -1), and its radius is 1.

Based on the given equation, you have correctly determined the center and the radius of the circle. It seems that you have done it correctly.