Answer:
2A) The centre is (-3,2) and the radius is 5.
Step-by-step explanation:
2A)
Okay so if you have the equation of a circle in standard from (ie it looks something like this:  (x-h)² + (y-k)² = r² ), it's pretty easy to find the centre and radius because in this form, r is the radius and (h,k) is the centre of the circle.
Note that (x-h)² + (y-k)² = x²+(-2h)x+h² + y²+(-2k)y+k²
In our case [x²+y²+6x-4y-12=0], we can see that -2h=6 (because 6 is the coefficient of x) and -2k=-4 (because -4 is the coefficient of y).
-2h=6, h=-3
-2k=-4, k=2
Let's sub these in:
x²+(-2h)x+h² + y²+(-2k)y+k² = r²
x²+6x+(-3)² + y²-4y+(2)² = r²
x²+6x+9 + y²-4y+4 = r²
x²+y²+6x-4y+(13-r²)=0
x²+y²+6x-4y+(13-r²)=x²+y²+6x-4y-12=0
13-r²=-12
r²=25
r=5
So our equation in standard form is (x+3)²+(y-2)²=25, which means the centre is (-3,2) and the radius is 5.
