hoopshiree one year ago what is the value of b x^2+y^2+ax+by+c=0

1. helder_edwin

the value of \(b\) for what?

2. hoopshiree

for that equation, \[x ^{2}+y ^{2^{}}+ax+by+c=0\]

3. hoopshiree

@helder_edwin

4. helder_edwin

r u sure the question is exactly that? it kind of doesn't make sense

5. hoopshiree

this graph is part of the question

6. helder_edwin

u also need to find the values of a, b, and c so that the equation describes the circle of the figure. complete both squares. u know the center and the radius of the circle (see the graph)

7. hoopshiree

that doesnt help me

8. hoopshiree

but thanks

9. helder_edwin

do u know how to complete the square?

10. hoopshiree

no

11. helder_edwin

ok i will do it. first the center is C(-2,1) and r=3, do u agree?

12. hoopshiree

yes

13. hoopshiree

I got that far.

14. helder_edwin

ok know, the equation of a circle should be \[\large (x-h)^2+(y-k)^2=r^2 \] we have to make the equation \[\large x^2+y^2ax+by+c=0 \] to resemble this. OK?

15. helder_edwin

\[\large x^2+y^2+ax+by+c=0 \]

16. hoopshiree

(x+2)^2+(y-1)^2=3^2

17. helder_edwin

we know that h=-2, k=1, and r=3, agree?

18. helder_edwin

yes. expand that!

19. helder_edwin

can u do it?

20. hoopshiree

(x+2)+(x+2)+(y-1)+(y-1)=9?

21. hoopshiree

I'm so confuseddd.

22. helder_edwin

no \[\large (x+2)^2+(y-1)^2=3^2 \] \[\large (x^2+4x+4)+(y^2-2x+1)=9 \] \[\large x^2+y^2+4x-2y+5=9 \] \[\large x^2+y^2+4x-2y-4=0 \]

23. helder_edwin

got it?

24. hoopshiree

so b is 4?

25. helder_edwin

no, b is the coefficient of \(y\)

26. hoopshiree

oh -2?

27. helder_edwin

yes

28. hoopshiree

can you help me with this one?

29. helder_edwin

same question?

30. hoopshiree

have to find a now. so the center is (-1,3) and radius is 4 right?

31. hoopshiree

(x+1)+(y-3)=16?

correction number one, the centre of circle is (1,-2) not (-2,1). so, h = 1 and k = -2 now remember the equation of circle is x^2 + y^2 +ax + by + c = 0 or (x - h)^2 + (y -k)^2 = 0 h,k is the centre of circle. the relation of b and k is b = -1/2(k) so, b = -1/2(-2) = 1

33. helder_edwin

YES. I am so sorry, my mistake.

34. hoopshiree

oh boy. okay.

35. helder_edwin

in the second case the center is (3,-1) and r=4

correction again, im not a boy, but a father of that boy ;p

37. hoopshiree

does A have a certain formula that can be used to find it?

38. helder_edwin

dont learn formulas, learn concepts. do what i did before: expand \[\large (x-3)^2+(y+1)^2=4^2 \] it should take a minute

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a = -1/2 * h

41. hoopshiree

so (x^2-6x+9)+(y^2+2x+1)=16?

42. hoopshiree

so a would be = -3 1/2

43. helder_edwin

a=-6, that is what u got.

44. ranga

It should be (x^2-6x+9)+(y^2+2y+1)=16 (you had 2x, it should be 2y). In bx^2+y^2+ax+by+c=0, a is the coefficient of x. The coefficient of x in the equation you got is -6 So a = -6.