Question

please help!! (its due very soon im keeping my fingers crossed i hope to get it over with ) The question will be graded based on mathematical content , reasoning skill, and communication and is worth 18 points

Answer 1

2A) The centre is (-3,2) and the radius is 5.

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.