anonymous
  • anonymous
what is the value of b x^2+y^2+ax+by+c=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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helder_edwin
  • helder_edwin
the value of \(b\) for what?
anonymous
  • anonymous
for that equation, \[x ^{2}+y ^{2^{}}+ax+by+c=0\]
anonymous
  • anonymous
@helder_edwin

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helder_edwin
  • helder_edwin
r u sure the question is exactly that? it kind of doesn't make sense
anonymous
  • anonymous
this graph is part of the question
helder_edwin
  • 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)
anonymous
  • anonymous
that doesnt help me
anonymous
  • anonymous
but thanks
helder_edwin
  • helder_edwin
do u know how to complete the square?
anonymous
  • anonymous
no
helder_edwin
  • helder_edwin
ok i will do it. first the center is C(-2,1) and r=3, do u agree?
anonymous
  • anonymous
yes
anonymous
  • anonymous
I got that far.
helder_edwin
  • 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?
helder_edwin
  • helder_edwin
\[\large x^2+y^2+ax+by+c=0 \]
anonymous
  • anonymous
(x+2)^2+(y-1)^2=3^2
helder_edwin
  • helder_edwin
we know that h=-2, k=1, and r=3, agree?
helder_edwin
  • helder_edwin
yes. expand that!
helder_edwin
  • helder_edwin
can u do it?
anonymous
  • anonymous
(x+2)+(x+2)+(y-1)+(y-1)=9?
anonymous
  • anonymous
I'm so confuseddd.
helder_edwin
  • helder_edwin
no \[\large (x+2)^2+(y-1)^2=3^2 \] \[\large (x^2+4x+4)+(y^2-2x+1)=9 \] \[\large x^2+y^2+4x-2y+5=9 \] \[\large x^2+y^2+4x-2y-4=0 \]
helder_edwin
  • helder_edwin
got it?
anonymous
  • anonymous
so b is 4?
helder_edwin
  • helder_edwin
no, b is the coefficient of \(y\)
anonymous
  • anonymous
oh -2?
helder_edwin
  • helder_edwin
yes
anonymous
  • anonymous
can you help me with this one?
helder_edwin
  • helder_edwin
same question?
anonymous
  • anonymous
have to find a now. so the center is (-1,3) and radius is 4 right?
anonymous
  • anonymous
(x+1)+(y-3)=16?
RadEn
  • RadEn
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
helder_edwin
  • helder_edwin
YES. I am so sorry, my mistake.
anonymous
  • anonymous
oh boy. okay.
helder_edwin
  • helder_edwin
in the second case the center is (3,-1) and r=4
RadEn
  • RadEn
correction again, im not a boy, but a father of that boy ;p
anonymous
  • anonymous
does A have a certain formula that can be used to find it?
helder_edwin
  • 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
SnuggieLad
  • SnuggieLad
Welcome To Openstudy
RadEn
  • RadEn
a = -1/2 * h
anonymous
  • anonymous
so (x^2-6x+9)+(y^2+2x+1)=16?
anonymous
  • anonymous
so a would be = -3 1/2
helder_edwin
  • helder_edwin
a=-6, that is what u got.
ranga
  • 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.

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