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

- anonymous

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

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- schrodinger

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- helder_edwin

the value of \(b\) for what?

- anonymous

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

- anonymous

@helder_edwin

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- helder_edwin

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

- anonymous

this graph is part of the question

##### 1 Attachment

- 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

that doesnt help me

- anonymous

but thanks

- helder_edwin

do u know how to complete the square?

- anonymous

no

- helder_edwin

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

- anonymous

yes

- anonymous

I got that far.

- 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

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

- anonymous

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

- helder_edwin

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

- helder_edwin

yes. expand that!

- helder_edwin

can u do it?

- anonymous

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

- anonymous

I'm so confuseddd.

- 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

got it?

- anonymous

so b is 4?

- helder_edwin

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

- anonymous

oh -2?

- helder_edwin

yes

- anonymous

can you help me with this one?

##### 1 Attachment

- helder_edwin

same question?

- anonymous

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

- anonymous

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

- 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

YES. I am so sorry, my mistake.

- anonymous

oh boy. okay.

- helder_edwin

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

- RadEn

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

- anonymous

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

- 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

Welcome To Openstudy

- RadEn

a = -1/2 * h

- anonymous

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

- anonymous

so a would be = -3 1/2

- helder_edwin

a=-6, that is what u got.

- 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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