SOLUTION: Hi, these are five questions which I need help in Write the general equation for the circle that passes through the points: (1, 7) (8, 6) (7, -1) Write the general equation f

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Question 1119612: Hi, these are five questions which I need help in
Write the general equation for the circle that passes through the points:
(1, 7)
(8, 6)
(7, -1)
Write the general equation for the circle that passes through the points (1, 1), (1, 3), and (9, 2).
Write the general equation for the circle that passes through the points (- 5, 0), (0, 4), and (2, 4).
Write the general equation for the circle that passes through the points:
(-1, 2)
(4, 2)
(- 3, 4)
Write the general equation for the circle that passes through the points:
(0, 0)
(6, 0)
(0, - 8)
Found 4 solutions by Alan3354, Boreal, solver91311, ikleyn:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Write the general equation for the circle that passes through the points:
(1, 7)
(8, 6)
(7, -1)
-----------------
You can use a matrix:
| x   y   x^2+y^2  1|
| 1   7     50     1|
| 8   6    100     1| = 0
| 7  -1     50     1|

----------
I have an Excel sheet that gives the result.
=============
Here's the "hard way:"
Label the points A(1,7), B(8,6) and C(7,-1)
Find the perpendicular bisectors of AB and BC.
The intersection of those 2 is the center of the circle.
---
The distance from the center to any point is the radius.



Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
For the last one
(x-a)^2+(y-b)^2=r^2
For the origin, x and y are 0 so a^2+b^2=r^2
For (6, 0) (6-a)^2+(0-b^2)=r^2 and 36-12a+a^2+b^2=r^2
But since r^2=a^2+b^2
36-12a+a^2+b^2=a^2+b^2
12a=36 and a=3
similarly, for (0, -8) we have a^2+64+16b+b^2=a^2+b^2 and -16b=-64 and b=-4
The center is at (3, -4) and the radius is sqrt((-3)^2+4^2)=5
(x-3)^2+(y+4)^2=25

Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!


Principles:

The perpendicular bisectors of any chords of a circle intersect in the center of the circle.

The slope of a perpendicular bisector is the negative reciprocal of the slope of the original line.

The distance from the center to any of the three given points is the radius.

Procedure:

Select any pair of the given points, and , and use the slope formula to calculate the slope of the line containing those two points.



Then calculate the negative reciprocal of this slope, i.e.

Use the midpoint formulas to calculate the coordinates of the midpoint, of the chord segment:





Derive the slope-intercept form of an equation of the line that has the negative reciprocal slope and passes through the calculated midpoint, thus:



Repeat the above process for a different pair of the given points.

Take the RHS of each of the derived equations and set them equal to each other. Solve for to obtain the -coordinate of the center of the desired circle. Substitute this value and calculate the -coordinate of the center of the circle. You now have determined the circle center,

Use the distance formula with any one of the given points and the center, to calculate the measure of the radius, . Since you will ultimately require only the square of the radius, you need not do the square root step in the distance formula.

Finally, construct the Standard Form:



And then expand the binomials and collect like terms to create the General Form as required.



Where and


John

My calculator said it, I believe it, that settles it
Answer by ikleyn(52797)   (Show Source): You can put this solution on YOUR website!
.
You may also want to look into the lesson
    - Find the standard equation of a circle
in this site,  where you will find many similar solved problems.

Your direct analogue and sample is  Problem 6  of that lesson.


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