## anonymous 4 years ago Center of a circle with this equation: y^2-6x-8y+20=-x^2

1. Mertsj

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

2. anonymous

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

3. anonymous

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

4. Mertsj

What negative?

5. anonymous

The neg in front of the twenty

6. Callisto

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?

7. Mertsj

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

8. Mertsj

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

9. anonymous

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

10. Callisto

No... the sign is not correct

11. anonymous

sooo it's (3,4).

12. anonymous

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

13. Callisto

Yes

14. anonymous

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

15. anonymous

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

16. Callisto

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

17. Callisto

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

18. anonymous

squareroot of 5...

19. Callisto

It should be that ...

20. anonymous

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

21. Callisto

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

22. Callisto

Here is the proof...