find the center (h,k) and radius r of the circle with the given equation:
x^2 + y^2 + 6x - 8y = 56
You must complete the square:
x^2 + 6x + y^2 - 8y = 56
For any (x-c), (x+c)^2 = x^2 + 2cx + c^2
Now compare x^2 + 6x to x^2 + 2cx + c^2. You are missing c^2. You can find c, though, by 6x:
6x = 2cx
6 = 2c
3 = c
So c^2 = 9.
The perfect square is (x+c)^2 = (x+3)^2 because you found that c=3
You do not have a 9 in the equation - but you can add 0:
x^2 + 6x + 9 - 9 + y^2 - 8y = 56
(x+3)^2 -9 + y^2 - 8y = 56
Move the 9 over to the other side
(x+3)^2 + y^2 - 8y = 64
Try doing the same with y.
To find the center (h,k) and radius r of a circle given its equation, we need to complete the square for both x and y.
1. First, group the x-terms and y-terms together:
x^2 + 6x + y^2 - 8y = 56
2. To complete the square for the x-terms, we need to add and subtract (6/2)^2 = 9 to both sides of the equation:
x^2 + 6x + 9 + y^2 - 8y = 56 + 9
Simplifying:
(x + 3)^2 + y^2 - 8y = 65
3. Now, let's complete the square for the y-terms. To do that, we need to add and subtract (8/2)^2 = 16 to both sides of the equation:
(x + 3)^2 + y^2 - 8y + 16 = 65 + 16
Simplifying:
(x + 3)^2 + (y - 4)^2 = 81
Now the equation is in the standard form of a circle: (x - h)^2 + (y - k)^2 = r^2, where (h,k) represents the center of the circle, and r represents the radius.
Comparing the given equation with the standard form, we can see that the center (h, k) is (-3, 4), and the radius r is the square root of 81, which is 9.
Therefore, the center of the circle is (-3, 4), and the radius is 9.