Question

Find the center and radius of the sphere with equation x^2 + 3x + y^2 -4y + z^2 +12z = 36

Answer 1

Okay... you need to master a little technique called... completing the square... you know of it? :D

Answer 2

Yes, somewhat.

Answer 3

\[x ^{2}+3x+___+y ^{2}-4y+___z ^{2}+12z+_____=36+_____+______+_____\]
fill in completing the squares.

Answer 4

Just in case you need a little review, here's an example of completing the square...
\[\Large w^2+\color{red}4w =32\]

the number in red (the one that is next to the variable WITH NO EXPONENT)
that's the number you focus on. What you do is
-GET HALF OF THAT NUMBER
-AND THEN SQUARE THAT.

To demonstrate,
-half of 4 is 2
-square of 2 is 4

So, what we do now is add 4 to BOTH SIDES of the equation, like so
\[\Large w^2 +4w \color{red}{+4}=32\color{red}{+4}\]

You'll notice that the left side is not what you'd call a perfect square trinomial, illustrating, let's factor it, we get...
\[\Large \color{blue}{(w+2)^2}=\color{red}{36}\]

And that's completing the square, in a nutshell
:D

Answer 5

I hate critical typos, I meant
"You'll notice that the left side is NOW what you'd call..."

Answer 6

9/4, 4, 36 and then all of those values are also added to other side

Answer 7

Impressive :)
\[\large \left(x^2+3x + \frac94 \right)+(y^2 - 4y +4)+(z^2+12z+36)=36+\frac94+4+36\]

Answer 8

You'll notice now that the left side now consists of three perfect square trinomials, right?

Answer 9

Yes.

Answer 10

I dont know how to do the first. the fraction messes me up

Answer 11

But the others should be (y+2)^2 + (z+6)^2

Answer 12

Okay, just a hint, remember the first thing you did with the coefficient of w?
Half of it, that would be the constant term in each binomial.
So with the x's, you had a 3/2 before you squared it... so...

\[\large \left(x+\frac32\right)^2+(y -2)^2+(z+6)^2=\frac{313}4\]

Answer 13

Catch me so far?

Answer 14

Okay, yes.

Answer 15

From here, it's just easy pickings.
The general equation of a sphere is
\[\Large (x-\color{red}h)^2 + (y -\color{blue}k)^2+(z-\color{green}l)^2 = \color{violet}r^2\]

If you have the equation in this form, then your center is simply
\[\Large (\color{red}h \ , \color{blue}k \ , \color{green}l)\]


And your radius is simply \(\Large \color{violet}r\)

Answer 16

So... can you now single out the values of h, k, l and r?

Answer 17

(-3/2,2,-6)

Answer 18

and radius should be Squareroot (313/4)

Answer 19

or sqrt(313) / 2

:)

Answer 20

Very nice though :)

I can see you just needed a nudge in the right direction, didn't you? ;)

Answer 21

Yes I did! Thank you for your help!!!

Answer 22

No problem :)

Answer 23

@terenzreignz good nudging.