Question

what is the center of the circle of this given equation? x2+y2+14x-12y=-69 Please show work so I can understand it. Thanks

Answer 1

there is a easy way do u know differentiation?

Answer 2

no

Answer 3

Group all the like variables together.
x^2 + 14x + y^2 - 12y = -69
"Finish the square"
What does this mean? It means that you add some number such that you can express the x terms in the form (x + a)^2

So for example, in our problem, we would add 49. How we get this number that completes the square is by taking the number in front x, diving it in half, and then squaring it.

Answer 4

I will continue my explanation, I only submitted it because I don't want you to wait 15 minutes and forget about it. Continuing on now.

Answer 5

so now we have:
x^2 + 14x + 49 + y^2 - 12y = -69

If we continue on from here, this statement would be invalid. Because we added 49 to 1 side, we must do so with the other:

x^2 + 14x + 49 + y^2 - 12y = -20

Just like we did for the x terms, we do so with the y terms.

So we take a look at the constant in front of y, which is -12. Half it, we get -6. Square it, we get 36. So we add 36 to both sides.

x^2 + 14x + 49 + y^2 -12y + 36 = 16
Now that we have completed the square for x and y, we can express it in the form:
(x + a)^2 + (y +b)^2 = some number

(x + 7)^2 + (y -6)^2 = 16
This is an equivalent statement. I'm assuming you know how to un-expand quadratic equations.

Now we set the x term being squared equal to 0, and solve for x, telling us the x-coordinate for the center. We do the same for y.

x + 7 = 0
So, x= - 7
y - 6 = 0
y = 6

So our center is at (-7, 6).

Answer 6

thank you so much!