## anonymous one year ago Please help!!!!! MEDAL AND FAN!!!!! Give the center coordinates (x,y) of the circle:

1. anonymous

$(x+3)^{2}+(y-4)^{2}=36$

2. triciaal

equation of a circle (x-h)^2 + (y-k)^2 = r^2 center (h,k) and radius r

3. anonymous

Center = (-3,4) This is because, as triciaal said, in the equation (x+3)2+(y−4)2=36, 3=h and 4=k (H,K) are the coordinates of the center.

4. anonymous

Oh ok

5. amilapsn

The definition of the circle is: the loci of a point which is moving at constant distance from a fixed point. So if the point is (h, k) and the constant distance is r, by the distance formula: $\sqrt{(x-h)^2+(y-k)^2}=r$ After squaring, equation of the circle: $(x-h)^2+(y-k)^2=r^2$

6. anonymous

When multiplied out, we obtain the "general form" of the equation of a circle. Notice that in this form we can clearly see that the equation of a circle has both x2 and y2 terms and these terms have the same coefficient (usually 1).

7. anonymous

Wait so which explanation should I go with because I a bit confused

8. anonymous

@ikram002p

9. triciaal

all are correct! different amount of details.

10. anonymous

ok thx

11. anonymous

yeah there all the same

12. anonymous

But..... the first time I did this problem, I put -3,4 for my answer and got it wrong

13. freckles

ordered pairs should be in the form (h,k) that is you should have (-3,4)

14. freckles

Maybe they already included the ( ). Don't know.

15. anonymous

oh ok I will put it in parentheses

16. anonymous

Could y'all help me on on more problem?

17. anonymous

*one

18. anonymous

Sure!

19. anonymous

$(x + 1)^{2} + (y - 6)^{2}= 2.25$

20. anonymous

Would the coords be (-1,6)?

21. freckles

yes the center would be (-1,6)

22. anonymous

Absolutly right!

23. anonymous

thx

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