Question

PLEASE HELP! :)
If x²+4x+y²-6y-12=0 is the equation of a circle, the length of the radius is

1. 25
2. 16
3. 5
4. 4

Answer 1

complete the square
get it to form
(x-h)²+(y-k)²=r²
r is radius
so
(x²+4x)+(y²-6y)-12=0
take 1/2 of each linear coefient and square it and add it to both sides
4/2=2, 2²=4
-6/2=-3, (-3)²=9
add 4+9 to both sides
(x²+4x+4)+(y²-6y+9)-12=9+4
factor
(x+2)²+(y-3)²-12=13
add 12 both sides
(x+2)²+(y-3)²=25
(x+2)²+(y-3)²=5²
radius is 5

3rd option

Answer 2

hello :
an equation of the circle Center at the A(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²
x²+4x+y²-6y-12=0
(x²+4x+2²) -2² +(y²-6y+3²)-3²-12 = 0
(x+2)² +(y-3)² = 25 = 5²  .....the length of the radius is : 5 (answer :3 )