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