Question

Help me set up this equation into a circle

Answer 1

Move all the x' s and y's to the left of the = sign

Answer 2

\[x^2+y^2+6x-8y+9=0\]

Answer 3

Leave the 9 where it is on the right....

Answer 4

Like legit the answer is (x+3)^2+(Y-4)^2=16. How do I get there?

Answer 5

One step at a time 9-).

So if you move all the x's and y's to the left , what do you get?

Answer 6

\[x^2+y^2+6x-8y+9=0\] I get this..

Answer 7

Ok. Lets take a step back. Start with the original equation, and subtract 6x from both sides

Answer 8

Hi I appreciate your help @fwizbang but \[x^2+y^2+6x-8y+9=0\] is the original equation. I just mistyped it

Answer 9

Ah, I understand. We have x^2 +6 x, what do we need to add to make this a perfect square?

Answer 10

I've tried (x^2+6x+9)+(y^2-8y) --> (x+3)^2 + (y^2-8y) and it's somewhat close to the answer : (x+3)^2+(Y-4)^2=16 but I can't get the (y^2-8y) part because I'm left with

Answer 11

(y^2-8y)

Answer 12

You're almost there. What do you need to add to y^2-8y=y^2+2*(-4)*y to make a perfect square?

Answer 13

I don't think its possible to factor down y^2-8y. Unless I do this y(y-8) which is unless if I need it to be y^2

Answer 14

Yes, but we can complete the square by adding a constant to both sides of the equation

Answer 15

Can you show me?

Answer 16

y^2 + 2*a*y = (y+a)^2 -a^2

Answer 17

So y^2-8*y= y^2-2(-4)y =?

Answer 18

yup

Answer 19

wait but wasnt the original -8y?? How can it be -2(-4y)?

Answer 20

whoops, y^2-8y=y^2+2(-4)y=?

Answer 21

so it would be (y-4)^2=4^2?

Answer 22

(y-4)^2+4^2

Answer 23

Is there an name for this rule?

Answer 24

-4^2. Then you add 4^2 to both sides to get the answer

Answer 25

Its called completing the square

Answer 26

Lol I got itt

Answer 27

Thanks for the help!! I really appreciate it!

Answer 28

general eq. of a circle is
\[x^2+y^2+2gx+2fy+c=0\]
where the center is (-g,-f) and radius
\[r=\sqrt{g^2+f^2-c}\]
to change this eq. in the form \[\left( x-h \right)^2+\left( y-k \right)^2=r^2\]
where center is (h,k) and radius r
\[\left( x^2+2gx \right)+\left( y^2+2fy \right)=-c\]
adding both sides
\[\left( x^2+2gx+g^2 \right)+\left( y^2+2fy+f^2 \right)=g^2+f^2-c\]
\[\left( x+g \right)^2+\left( y+f \right)^2=\left( \sqrt{g^2+f^2-c} \right)^2\]
which is of the form \[\left( x-h \right)^2+\left( y-k \right)^2=r^2\]
where h=-g.k=-f,\[r=\sqrt{g^2+f^2-c}\]