## iheartfood 2 years ago What is the value of a for the attached circle in general form?

1. iheartfood

2. iheartfood

**no answer choices!! and idk how to solve :(

3. UnkleRhaukus

factor the equation using prefect squares

4. iheartfood

which equation??

5. iheartfood

like the one of the circle?? (which i find??) or the general form?

6. UnkleRhaukus

$x^2+ax+y^2+by+c=0$ $\left(x+\frac a2\right)^2-a^2+y^2+by+c=0$

7. iheartfood

why whoops?? haha :P am i not seeing something ur seeing??

8. UnkleRhaukus

(i forgot to square ) (fixed now)

9. iheartfood

ohhh okay i see :P so what do i do now?

10. UnkleRhaukus

factor the y terms using the same technique , then move the three constants outside the brackets to the right hand side

11. iheartfood

(x+a/x)^2=(-a+y+by)^2 ?? did i do that right?? :/ idk...

12. iheartfood

oops i forgot the +c... so like this?? (x+a/x)^2=(-a+y+by)^2+c ??? idk :(

13. UnkleRhaukus

$\left(x+\frac a2\right)^2-a^2+y^2+by+c=0$ $\left(x+\frac a2\right)^2-a^2+\left(y+\frac b2\right)^2-b^2+c=0$ $\left(x+\frac a2\right)^2+\left(y+\frac b2\right)^2=a^2+b^2-c$

14. iheartfood

oh okay :P but how do i solve for a?

15. UnkleRhaukus

compair to the general form of a circle

16. iheartfood

|dw:1354600365624:dw|

17. iheartfood

or the one in my question?

18. UnkleRhaukus

|dw:1354600399190:dw|

19. iheartfood

oh sorry! i meant - haha :P so do i refer to that one?? or the one in my problem? this one?|dw:1354600447714:dw|

20. UnkleRhaukus

|dw:1354600492668:dw|

21. UnkleRhaukus

now look at the graph

22. iheartfood

okay so (h,k) center=(-4,2) and radius=2 ???

23. iheartfood

is that right?? but how do i get the value of a from that?

24. UnkleRhaukus

$\left(x+\frac a2\right)^2+\left(y+\frac b2\right)^2=a^2+b^2-c$ $\left(x-h\right)^2+\left(y-k\right)^2=r^2$

25. iheartfood

so a= -4/2 ???

26. iheartfood

so a=-2 ?? is that right?? my answer is -2??

27. jim_thompson5910

center=(-4,2) ---> h = -4, k = 2 radius=2 ---> r = 2 (x-h)^2 + (y-k)^2 = r^2 (x-(-4))^2 + (y-2)^2 = 2^2 (x+4)^2 + (y-2)^2 = 4 Now get this equation into x^2 + y^2 + ax + by + c = 0 form

28. iheartfood

okay, lemme give this a shot hahaha idk how close ill get tho!!! x^2 + y^2 + 4^2 - 2^2 - 4 = 0 ??? idk haha :P

29. iheartfood

is that right?? or even close?? :/

30. jim_thompson5910

no that's not correct

31. jim_thompson5910

(x-h)^2 + (y-k)^2 = r^2 (x-(-4))^2 + (y-2)^2 = 2^2 (x+4)^2 + (y-2)^2 = 4 x^2 + 8x + 16 + y^2 - 4y + 16 = 4

32. jim_thompson5910

keep going

33. iheartfood

aww darn :( okay so i simplify that right?

34. jim_thompson5910

yes

35. iheartfood

kk so i get this? x^2 + 8x + 16 + y^2 - 4y + 16 = 4 = x^2 + y^2 +8x -4y +12 = 0 ??

36. jim_thompson5910

not quite

37. jim_thompson5910

16+16-4 isn't 12

38. iheartfood

ohhh theres another 16!! i didn't see that! my bad!!! so i get this?? x^2 + y^2 +8x -4y + 28 = 0

39. jim_thompson5910

yep

40. iheartfood

kk so a=8 then?? my answer is 8? :O

41. jim_thompson5910

yes

42. iheartfood

awesome :) thanks so much for all of ur help @UnkleRhaukus and @jim_thompson5910 :)