Question

How do I algebraically complete the square of these equaitons?

Answer 1

\[AX^2+CX+E=0\]

Answer 2

you mean
\[(AX^2+CX+E)^2=0^2\]
?

Answer 3

What I am trying to find is the algebraic equation of a circle with non-one coefficients of x^2 and y^2. A and B cannot be one in the following equation.

\[AX^2+BY^2+CX+DY+E+0\]

Answer 4

Excuse me. That was =0 not +0.

Answer 5

ohh ok... sorry i misread

Answer 6

So what I am assuming is that I break it down into the x and y gorups.\[(AX^2+CX)+(BY^2+DY)+E\]

Answer 7

i dont think you will get a circle... you will get an ellipse

Answer 8

So long as A+B, it is a circle.

Answer 9

A=B*

Answer 10

Ultimately, I am trying to find an algebraic expression that I can use to find the radius and center of the circle.

Answer 11

Then, I divide each side by the coefficient of ^2.
\[X^2+(CX/A)+Y^2+(DY/B)+E=0\]

Answer 12

\[X^2/B+(CX/A)+Y^2/A+(DY/B)+E=0\]

thats what you get if you divide throughout

Answer 13

\[X^2/B+(CX/AB)+Y^2/A+(DY/AB)+E/AB=0\]
actually this

Answer 14

Okay that makes sense. Howe do I further reduce that equation?

Answer 15

My teacher made it sound like we would be completing the square, but I do not know how to do that twice in this sense.

Answer 16

its easier to work with
\[(AX^2+CX)+(BY^2+DY)+E\]

Answer 17

He wants it to be in the standard form of a circle, so that we can easily find (h,k) and r.

Answer 18

wait, let me give it a try on paper
but i dont think you will get a circle

Answer 19

You do get a circle, only if A+B.

Answer 20

\[(AX^2+CX+C^2/4A)-C^2/4A+(BY^2+DY)+E=0 \\ (\sqrt{A}x + C/2\sqrt{A})^2+(BY^2+DY)+E=-C^2/4A\]

Answer 21

i've completed the square for the 'x' terms
you have to do the same thing for the y terms

Answer 22

Haha Thanks so much man.

Answer 23

It is exactly the same thing for y correct? I will just replace y for x and B for A etc etc.

Answer 24

yep

Answer 25

Awesome

Answer 26

How do I combine both of those into one equation then?

Answer 27

have you been told A=B?
then we could have just set A=B and worked it so that you get a circle...

Answer 28

because working the equation i just got into an ellipse looks very very tedious

Answer 29

Yes A=B

Answer 30

ok then,
lets re do with this
\[X^2/B+(CX/AB)+Y^2/A+(DY/AB)+E/AB=0\]

Answer 31

woops
i mean this\[AX^2+BY^2+CX+DY+E+0\]

Answer 32

but A=B
so
\[AX^2+AY^2+CX+DY+E+0\]

Answer 33

Back at square one lol

Answer 34

hehe yep...
divide through out by A
\[X^2+Y^2+CX/A+DY/A+E/A+0\]

Answer 35

now lets make these substitutions for ease

let C/A=L
D/A=M
E/A=N

so we have
\[X^2+Y^2+LX+MY+N=0\]

Answer 36

yea, give me a sec to figure out what to add and subtract

Answer 37

Also, any tips for when A does not equal B.

Answer 38

\[(X^2+LX+L^2/4)+Y^2+MY+N=L^2/4 \\(X + L/2)^2+Y^2+MY+N=L^2/4 \]

Answer 39

repeat the same for Y

Answer 40

\[(X + L/2)^2+(Y + M/2)^2=L^2/4+M^2/4-N\]

Answer 41

Okay

Answer 42

for an ellipse, you would have to start this way
\[(AX^2+CX)+(BY^2+DY)+E =0\\ A(X^2+CX/A)+B(Y^2+DY/B)+E\]

Answer 43

All Right. Thank you so much for your help.

Answer 44

sure :)

Answer 45

Is there some way I rate you on hear or anything?

Answer 46

nope...
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