SOLUTION: Solve the exercise by solving a system of equations. Find the center and radius of the circle whose graph passes through the points (−5, −1), (−11, &#87

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Question 1060075: Solve the exercise by solving a system of equations.
Find the center and radius of the circle whose graph passes through the points
(−5, −1),

(−11, −9),
and
(−4, −8).
(Hint: Use the equation
x2 + y2 + ax + by + c = 0.)
Found 3 solutions by josgarithmetic, MathTherapy, Alan3354:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2By%5E2%2Bax%2Bby%2Bc=0
ax%2Bby%2Bc=-x%5E2-y%5E2------the equations of your system will use this format.

Each given point makes an equation for the system.

-5a-b%2Bc=-%2825%2B1%29
-5a-b%2Bc=-26
5a%2Bb-c=26
-
-11a-9b%2Bc=-121-81
-11a-9b%2Bc=-202
11a%2B9b-c=202
-
-4a-8b%2Bc=-%2816%2B64%29
-4a-8b%2Bc=-80
4a%2B8b-c=80
-
-
Solve this system for a, b, c:
system%285a%2Bb-c=26%2C11a%2B9b-c=202%2C4a%2B8b-c=80%29

Solve the first of these for c, and substitute into the other two equations and get simpler system system%286a%2B8b=176%2C-a%2B7b=56%29;

Equivalently this allows elimination of a:
system%286a%2B8b=176%2C-6a%2B4b=336%29;
-
highlight%28b=42%262%2F3%29

The first equation in the last system is same as 3a%2B4b=82.
Use b to find a.
3a=82-4b
3a=82-4%28128%2F3%29
a=%2882-4%28128%2F3%29%29%2F3
highlight%28a=-29%265%2F9%29

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the exercise by solving a system of equations.
Find the center and radius of the circle whose graph passes through the points
(−5, −1),

(−11, −9),
and
(−4, −8).
(Hint: Use the equation
x2 + y2 + ax + by + c = 0.)
It's beyond me why you're told to use the STANDARD form of the equation of a circle to find the center and radius. It's TOTALLY RIDICULOUS and only leads 

to a lot more work. When you use that form, and you determine, a, b, and c, you still have to complete the square in order to determine the center and radius.

This makes absolutely no sense. Anyway, if that's the way you have to do it, then do so. You should get: 



I recommend, however, that you use the center-radius form of the equation of a circle: %28x+-+h%29%5E2+%2B+%28y+-+k%29%5E2+=+r%5E2. It'll be much, much easier for you.

Using this form, you should get: , or the equation: highlight_green%28%28x+%2B+8%29%5E2+%2B+%28y+%2B+5%29%5E2+=+25%29%29.

I hope you IGNORED the RUBBISH (GARBAGE) the other person posted!

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
I made an Excel sheet that does that.
----
|x  y  x^2+y^2 1|
|x1 y1 x1+y1^2 1|
|x2 y2 x2+y2^2 1| = 0
|x3 y3 x3+y3^2 1|

------------
Without the sheet, I would do it like this:
A(-5,-1)
B(-11,-9)
C(-4,-8)
---
Find the perpendicular bisector of AB, and of BC
Find the intersection of the 2 bisectors, that's the center.
The radius is the distance from the center to any point.