Question

Using the following equation, find the center and radius:

x2 − 4x + y2 + 8y = −4

Answer 1

@whpalmer4

Answer 2

okay, you need to complete the square on both x and y to get it in form

(x-h)^2 + (y-k)^2 = r^2

and then r will be the radius, center at (h,k)

Answer 3

ok so what will I haft to do next

Answer 4

we have x^2-4x as terms in x, so we take half of the x term's coefficient, square it, and add that to both sides, giving us

[x^2 - 4x + (-4/2)^2] + y^2 + 8y = -4 + (-4/2)^2
[x^2 - 4x + 4] + y^2 + 8y = -4 + 4
now we can rewrite the x portion as (x-2)^2 because (x-2)^2 = x^2 - 4x + 4

(x-2)^2 + y^2 + 8y = 0

now you repeat the process on the y portion

Answer 5

how would I repeat it? please help I am very confused

Answer 6

(x-2)^2 + [y^2 + 8y] = 0
we want to rewrite the portion in the brackets as (y-k)^2 for some value of k

to do that, we take half of the value of the coefficient of the y term (not the y^2 term), square it, and add it to both sides of the equation (thus preserving the equation's correctness, but putting it in a more convenient form for us

remember

(y-k)(y-k) = y^2 -ky -ky + k^2 = y^2 -2ky + k^2

we have 8y and we need it to equal -2ky

8y = -2ky
divide both sides by y
8 = -2k
divide both sides by -2
-4 = k

so we can write (y-k)^2 = (y-4)^2 = y^2 - 4y - 4y + 16 = y^2 - 8y + 16

the only problem is we don't have a 16 floating around in our left hand side of the equation, so we have to create one by adding 16 to both sides.

(x-2)^2 + [y^2 + 8y + 16 ] = 0 + 16
now we can rewrite the bracketed portion as (y+4)^2 because (y+4)^2 = y^2 +8y +16

(x-2)^2 + (y+4)^2 = 16

Answer 7

now let's compare that with our standard formula for the circle:
(x-h)^2 + (y-k)^2 = r^2

the two are equal if h = 2, k = -4, r = sqrt(16) = 4

agreed?

Answer 8

so that means the center is at (h,k) = (2, -4)
the radius is r = 4

Answer 9

so will the answer be thisThe center is located at (2, −4), and the radius is 4