Verified answer

The equation of a circle centered in (h,k) and radius = r is :

(x-h)^2 + (y-k)^2 =r^2

so :

x^2 +h^2 -2*x*h +y^2 +k^2 -2*y*k -r^2= 0

multiply by 4 so that you can compare this equation with yours :

4x^2 +4*h^2 -8*x*h + 4y^2 +4*k^2 -8*y*k -4*r^2 =0

so :

-8*x*h=0 then h=0

-8*y*k= 4*y then k= -1/2

4*k^2-4*r^2= -15 then r=2

:)

A circle with center (a, b) and radius r is the set of all points (x, y) where r² = (x - a)² + (y - b)²

So complete the square for both x & y

4x^2+4y^2+4y−15=0

4 ( x^2 + y^2 + y) = 15

Divide both sides by 4

( x^2 + y^2 + y) = 15/4

x^2 + y^2 + y + 1/4 = 15/4 + 1/4

x^2 + (y+1/2)^2 = 4

x^2 + (y+1/2)^2 = 2^2

Hence H = 0 , K = -1/2 & r = 2

Thus your center is ( 0,-1/2 ) & radius = 2

You need to place the equation in the form (x+h)^2 + (y+k)^2 = R^2 where the center of the circle is at -h and -k and R is the radius of the circle.

FIrst divide through by 4 to get x^2+y^2+y=15/4

Since the x^2 term is the only term with the variable x, h would be zero.

For the y terms, complete the square of the y^2+y=15/4 to get y^2+y+1/4=15/4+1/4

which can be factored to (y+1/2)^2 = 16/4

So the final form of the circle equation is (x+0)^2 + (y+1/2)^2 = 4

The center of the circle is located at (0,-1/2) with radius 2