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

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

Mathematics
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Get the variables on the left and the constant term on the right. complete the square twice and write the equation in the form: \[(x-h)^2+(y-k)^2=r^2\]
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
I am confused as to what to do with the neg.

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What negative?
The neg in front of the twenty
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?
Nothing. It will be gone because you will be adding numbers to the right and the left sides.
Ok. callisto wants to help you now. Good bye and good luck.
center is therefore.. (-3,-4) right?
No... the sign is not correct
sooo it's (3,4).
So basically it's still the same center as it would be without the negatives...?
Yes
Okay. Also, what the radius would be sqrtroot of 20.. right?
Sorry, but just for conformation.. It doesn't matter if there is a negative for the 'r', the center would still be the same.
Nope.. \[r = \sqrt { (D/2)^2 + (E/2)^2 -F}\]
D=-6, E=-8 , F=20 in your case. Put the numbers into it and solve, what would you get for r?
squareroot of 5...
It should be that ...
That.. doesn't make sense though. The radius is typically sqroot of \[(x-h)^2 + (y-k)^2=r^2\]
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
Here is the proof...
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