SOLUTION: Find the equation of the circle passing through the points A(3,2),B(5,3) and C(2,5)

Algebra ->  Length-and-distance -> SOLUTION: Find the equation of the circle passing through the points A(3,2),B(5,3) and C(2,5)      Log On


   



Question 990592: Find the equation of the circle passing through the points A(3,2),B(5,3) and C(2,5)
Found 2 solutions by anand429, Edwin McCravy:
Answer by anand429(138) About Me  (Show Source):
You can put this solution on YOUR website!
Let centre be (x,y) and radius be r
Now ,
%28x-3%29%5E2+%2B+%28y-2%29%5E2+=+r%5E2+ ---------(i)
%28x-5%29%5E2+%2B+%28y-3%29%5E2+=+r%5E2+ ---------(ii)
%28x-2%29%5E2+%2B+%28y-5%29%5E2+=+r%5E2+ ---------(iii)
Subtracting (ii) from (i), we get,
4x+%2B+2y+-+21+=+0---------------------(iv)
Subtracting (iii) from (i), we get,
-2x+%2B+6y+-+16+=+0 --------------------(v)
Solving (iv) and (v), we get,
x = 47/14
y = 53/14
Put it in any equation to get equation of circle.
(book by S.L.Loney)

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
The easiest way is to use the fact that the perpendicular bisector of a chord
is an extended diameter. Then two perpendicular bisectors of chords must
intersect at the center of the circle.  Then since to have the center you will
be able to find the radius with the distance formula.

1. Find the slope of chord AB
2. Find the midpoint of chord AB
3. Find the equation of the perpendicular bisector of chord AB.
   (a) Use slope which is the negative reciprocal of the slope of AB,
   (b) Use midpoint of AB, 
   (c) Use the point-slope formula. 

4. Find the slope of either chord BC [or AC].
5. Find the midpoint of chord BC [or AC]
6. Find the equation of the perpendicular bisector of chord BC [or AC]
7. Solve the system of two equations obtained in steps 3 and 6.
8. The solution to 7 will be the center, (h,k)
9. Use the distance formula to find r, the radius which is the distance 
   from the center to any of the three given points/
10. Then substitute h, k, and r in this general form for the equation of a circle.:

      (x-h)2 + (y-k)2 = r2   

The other way, which is harder, is to substitute the three points into that
equation and get a system of three equations in h, k and r, and solve for those
variables.

      (x-h)2 + (y-k)2 = r2

Edwin