Question

x^2+y^2-8x+10y+15=0
I need to convert this to standard form, but I'm not quite sure how. . .

Answer 1

OK, and what should standard form look like when we are done? What does "standard form" mean?

Answer 2

ax^2+bx+c
I'm not sure how to answer what it means... It means like... it's in the standard form? .-.

Answer 3

No worries you answered just fine. :)

Actually the standard form you just gave is the standard form of a parabola (you use this when you have only x's). But we have both x's and y's.

So we need to use a different standard form. I'll give it to you, but does that make sense so far?

Answer 4

you need to convert it to the standard form of a circle

Answer 5

Yea, it does, I knew there were different forms of it, but I wasn't sure which one...

Answer 6

Right. So, since we have positive x^2 and positive y^2 in our equation, we will use try to make it look like the standard form of a circle.

(x - a)^2 + (y - b)^2 = r^2

Answer 7

I thought standard form was Ax + By = C ;o

Answer 8

There are many "standard forms" depending on what shape your equation represents.

Answer 9

Line: Ax + By = C
Parabola: Ax^2 + Bx + C = 0
Circle: (x - A)^2 + (y - B)^2 = R^2
and so on...

Answer 10

Of course, the question is "How do we get the mess we have into standard form?" :)

Answer 11

Let's look at what we are given:

x^2 + y^2 - 8x + 10y + 15 = 0

Answer 12

Here's the first part of the trick. We move the +15 (just a number) to the right side of the equation like this...

x^2 + y^2 - 8x + 10y = -15

Answer 13

Is that OK so far?

Answer 14

Yep c:
That's the one step I got and didn't know what to do next. n_n

Answer 15

Cool. Now, we need to be a little sneaky... I am going to pair the x's and y's together so that it looks more organized like this...

x^2 - 8x + y^2 + 10y = -15

Answer 16

Ok c:

Answer 17

OK, now we will imagine how we can factor the x's and the y's... like this

x^2 - 8x + y^2 + 10y = -15
(x )^2 + (y )^2 = -15

There's more just wait one sec. :)

Answer 18

To get a -8x, I need to put a -4 in the factor like this

x^2 - 8x + y^2 + 10y = -15
(x - 4 )^2 + (y )^2 = -15

Answer 19

If yhur trying to get a -8, and yhu put a -4 in and square it... won't yhu end up with a positive?

Answer 20

Aha! you are right, we have to be careful... notice that
(x - 4)^2 = (x - 4)(x - 4) = x^2 - 8x + 16 but we don't have a 16 to factor before!

So, I need to add this 16 to the original equation... It sounds complicated but it just looks like this:

x^2 - 8x + 16 + y^2 + 10y = -15 + 16
(x - 4)^2 + (y )^2 = -15 + 16

Answer 21

I haven't done anything wrong. I just added 16 to both sides of the equation which is OK to do. Kind of scary, but does it make sense so far? We are almost done. :)

Answer 22

Yes it does. n_n

Answer 23

OK, now the x's are all fixed. We only have to worry about the y's.

Answer 24

But we can just use the same trick as before! :D

Answer 25

I notice that if I put (y + 5)^2, I would get y^2 + 10y + 25...

Answer 26

I really like the y^2 + 10y part, and I can fix the + 25 part easily, like this...

Answer 27

x^2 - 8x + 16 + y^2 + 10y + 25 = -15 + 16 + 25
(x - 4)^2 + (y + 5)^2 = -15 + 16 + 25

Answer 28

We got it! All that remains is to add the numbers on the right together to get the final answer.

Answer 29

(x - 4)^2 + (y + 5)^2 = 26

The standard form is found! Whew... any questions?

Answer 30

:O No. Thank yhu for walking me through that. That was really helpful. n_n
I have one more like this, do yhu mind if i try it out and post it in a new thread and tag yhu?

Answer 31

Of course, let me know what you get. :)

Answer 32

Alright. n_n Thank yhu again

Answer 33

No problem!