Question

Center of a circle with this equation:
y^2-6x-8y+20=-x^2

Answer 1

Get the variables on the left and the constant term on the right. complete the square twice and write the equation in the form:
\[(x-h)^2+(y-k)^2=r^2\]

Answer 2

I did but I end up with what I believe is the wrong answer.
Okay, so obviously it would be (x-4)^2+(y-3)^3=-20

Answer 3

I am confused as to what to do with the neg.

Answer 4

What negative?

Answer 5

The neg in front of the twenty

Answer 6

y^2-6x-8y+20=-x^2
x^2 +y^2-6x-8y+20= 0
centre = (-D/2 , -E/2) in the form x^2 +y^2+Dx+Ey+F= 0
Therefore centre = ( -(-6/2) , -(-8/2) ) =......
Can you do it now?

Answer 7

Nothing. It will be gone because you will be adding numbers to the right and the left sides.

Answer 8

Ok. callisto wants to help you now. Good bye and good luck.

Answer 9

center is therefore.. (-3,-4) right?

Answer 10

No... the sign is not correct

Answer 11

sooo it's (3,4).

Answer 12

So basically it's still the same center as it would be without the negatives...?

Answer 13

Yes

Answer 14

Okay. Also, what the radius would be sqrtroot of 20.. right?

Answer 15

Sorry, but just for conformation.. It doesn't matter if there is a negative for the 'r', the center would still be the same.

Answer 16

Nope..
\[r = \sqrt { (D/2)^2 + (E/2)^2 -F}\]

Answer 17

D=-6, E=-8 , F=20 in your case.
Put the numbers into it and solve, what would you get for r?

Answer 18

squareroot of 5...

Answer 19

It should be that ...

Answer 20

That.. doesn't make sense though. The radius is typically sqroot of \[(x-h)^2 + (y-k)^2=r^2\]

Answer 21

So, let's do the completing square once
y^2-6x-8y+20=-x^2
x^2 + y^2-6x-8y+20= 0
x^2 -6x +y^2-8y+20 =0
(x^2 -6x +9 -9) +(y^2-8y +16 -16)+20 =0
(x^2 -6x +9) +(y^2-8y +16) -9 -16 +20 =0
(x-3)^2 + (y-4)^2 - 5 =0
(x-3)^2 + (y-4)^2 = 5
(x-3)^2 + (y-4)^2 =[ sqrt (5) ]^2
r = sqrt 5
It DOES make sense

Answer 22

Here is the proof...