Question

PLEASE HELP! WILL FAN AND MEDAL! :):)

Using the following equation, find the center and radius of the circle. You must show all work and calculations to receive credit.

x2 − 4x + y2 + 8y = −4

Answer 1

lol u do it it is easy

Answer 2

@jim_thompson5910 sorry.. i didn't realize i closed it

Answer 3

this is all i got...

The general equation of a circle is:
x2+y2+2gx+2fy+c=0....where the center is:
(−g,−f)

so if you subtract 4 from both sides of the equation you get:
x2+y2−4x+8y+4=0

2g is -4 and 2f is 8
so the center is (-g,-f) which is (2,-4)

Answer 4

i remember most of it.. but i know im missing something in the equation..

Answer 5

if we just focus on x^2-4x, what do we need to complete the square?

Answer 6

example:

say we had x^2 + 12x

to complete the square, we take half of 12 (the x coefficient) to get 6
then we square 6 to get 36. This is added and subtracted onto x^2+12x to get `x^2+12x+36-36`

notice how the `x^2+12x+36` factors to `(x+6)^2`

so `x^2+12x` turns into `x^2+12x+36-36` which becomes `(x+6)^2-36`

Answer 7

oh.. i think ive been doing this all wrong...

Answer 8

i did... (x² − 4x) + (y² + 8y) = −4
(x² - 4x + 2²) + (y² + 8y + 4²) = -4 + 2² + 4²
(x-2)² + (y+4)² = 4²
center (2,-4)

Answer 9

What you posted above is correct, but this is how I'm familiar with this type of problem.

Answer 10

ok one sec

Answer 11

you have the correct steps, that leads to the radius being r = 4

Answer 12

you also have the correct center

Answer 13

oh.. well.. how did you get 4?

Answer 14

see attached

Answer 15

Your last equation is in the form
\[\Large (x-h)^2 + (y-k)^2 = r^2\]
the right side, which is 4^2 is equal to r^2

r^2 = 4^2
r = 4

Answer 16

oh wow.. now i feel really stupid haha

Answer 17

nah, you just overlooked it

Answer 18

thank you for your help! i really appreciate it :)

Answer 19

you're welcome