PLEASE HELP I AM IN DYER NEED!

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rachael724

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Find a,b, and c so that the graph of the circle with equation x^2 + y^2 + ax + by + c = 0 passes through with the points (-4,1), (-1,2), and (3,-6).

1 =
2 =
-6 =


Now I am stuck on how to start this because it confuses me because we mhave a, b and c as well as x and y. I need to show my steps and don't know how to do my first step.
 
Substitute the three sets of values for x and y, then solve for a, b, and c.

x^2 + y^2 + ax + by + c = 0

(-1,2) ==> x = -1, y = 2

(-1)^2 + (2)^2 + a(-1) + b(2) + c = 0

etc.
 
is this right so far?

1 + 4 + (-1)a + 2b + c = 0

5 + (-1)a + 2b + c = 0
-5 -5

(-1)a + 2b + c = -5

This isn't looking right....... ahhhhh help!
 
x^2 + y^2 + ax + by + c = 0
(-4,1) ==> x = -4, y = 1
(-4)^2 + (1)^2 + a(-4) + b(1) + c = 0
16 + 1 + (-4)a + b + c = 0
(-4)a + b + c = -17

x^2 + y^2 + ax + by + c = 0
(3,-6) ==> x = 3, y = -6
(3)^2 + (-6)^2 + a(3) + b(-6) + c = 0
9 + (-36) + 3a - (-6)b + c = 0
-27 + 3a + 6b + c = 0
3a + 6b + c = 27
 
Better check your math....


x^2 + y^2 + ax + by + c = 0
(3,-6) ==> x = 3, y = -6
(3)^2 + (-6)^2 + a(3) + b(-6) + c = 0 right
9 + (-36) + 3a - (-6)b + c = 0 what does (-6)^2=?
-27 + 3a + 6b + c = 0
3a + 6b + c = 27
 
x^2 + y^2 + ax + by + c = 0
(3,-6) ==> x = 3, y = -6
(3)^2 + (-6)^2 + a(3) + b(-6) + c = 0
9 + (36) + 3a - (-6)b + c = 0
45 + 3a + 6b + c = 0
3a + 6b + c = - 45

now what?
 
rachael724 said:
now what?
Click to expand...
Once you have a system of three linear equation in three variables ("a", "b", and "c"), solve the system by whatever method (matrices, Cramer's Rule, etc) that you prefer.

Then write the circle equation with the values you've found.

Eliz.
 
rachael724 said:
What method is easier?
Click to expand...
I don't know that one method is always better than another, though matrices would probably be quicker than Cramer's Rule in this particular case (since you need all three values, not just one), and more so if you have a calculator that can do the matrix for you.

But don't go by my opinion. Just choose whichever method you like. :D

rachael724 said:
Can you give me an example like before.
Click to expand...
I don't recollect the thread in which I gave a worked example of solving a system of linear equations. Please forgive my forgetfulness and reply with a link. Thank you.

Eliz.
 
It is hoped that at least sometimes all you need it is a little shove in the right direction. :)
 
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