SOLUTION: Find the equation of the circle in the form X^2+y^2+Dx+Ey+F=0 of the circle through points (1,1) (2,-1), (2,3)

Algebra ->  Circles -> SOLUTION: Find the equation of the circle in the form X^2+y^2+Dx+Ey+F=0 of the circle through points (1,1) (2,-1), (2,3)       Log On


   



Question 771638: Find the equation of the circle in the form X^2+y^2+Dx+Ey+F=0 of the circle through points (1,1) (2,-1), (2,3)

Found 2 solutions by josgarithmetic, MathLover1:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
The actual variables will be D, E, and F. The x and y values will become constants or part of constants. The system of equations will be more comfortable with each equation in the form, xD%2ByE%2BF=-x%5E2-y%5E2.

Each point gives one equation:
SYSTEM:
1D%2B1E%2BF=-1-1
2D-1E%2BF=-2%5E2-%28-1%29%5E2
2D%2B3E%2BF=-2%5E2-3%5E2
'
SIMPLIFIED SYSTEM:
---------------------------
D%2BE%2BF=-2
2D-E%2BF=-5
2D%2B3E%2BF=-13
---------------------------

You can solve for the three variables any way you know or like. You can also solve using the matrix,



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Intentionally unfinished allowing you to finish yourself.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Consider the general equation of a circle as X%5E2+%2B+Y%5E2+%2B+DX+%2B+EY+%2B+F+=+0.
given: points (1,1) (2,-1), (2,3)
Using the given three points we derive our equation :
From the point (1,1) we get the equation
1%2B+1+%2B+D+%2B+E+%2B+F+=+0 or
D+%2B+E+%2B+F%2B+2+=+0 ........ equation 1
From the point (2,-1) we get
4+%2B+1+%2B+2D+-E+%2B+F+=+0 or
2D+-E+%2B+F%2B5+=+0........ equation 2

From the point (2,3) we get
4+%2B+9+%2B2D+%2B+3E+%2B+F+=+0 or
2D+%2B+3E+%2B+F%2B13+=+0....... equation 3

We now have a system of three equations :

D+%2B+E+%2B+F%2B+2+=+0 ........ equation 1
2D+-E+%2B+F%2B5+=+0........ equation 2
2D+%2B+3E+%2B+F%2B13+=+0....... equation 3

We now use elimination method to find the value of D,+E and F.
Using equation 1 and eq. 2 to eliminate F, we do this by subtracting equation 2 from eq. 1 :
D+%2B+E+%2B+F%2B+2-%28+2D+-E+%2B+F%2B5+%29=+0
D+%2B+E+%2B+F%2B+2-2D+%2BE+-+F-5+=+0
2E+-2D-3+=+0..... let this be equation 4

Using Equation 2 and eqn 3, if we subtract eq. 3 from eq. 2 we can eliminate F and D:
2D+-E+%2B+F%2B5+-%282D+%2B+3E+%2B+F%2B13+%29=+0
2D+-E+%2B+F%2B5+-2D-3E+-+F-13+=+0
-E+%2B5+-3E+-13+=+0
-4E++-8+=+0
-E++-2+=+0 => highlight%28E+=-2%29
go back to eq.4 and plug in E
2%28-2%29+-2D-3+=+0.......solve for D
-4+-3+=+2D
-7+=+2D
-7+%2F2=+D
highlight%28D=-3.5%29
go back to eq.1 and plug in E and D and solve for F
-3.5-2+%2B+F%2B+2+=+0 ........ equation 1
-3.5+%2B+F+=+0
highlight%28F+=+3.5%29
We get this general equation by substituting
D+=+-3.5, E+=+-2 and F+=+3.5 to the general equation of the circle
X%5E2+%2B+Y%5E2+%2B+DX+%2B+EY+%2B+F+=+0:
X%5E2+%2B+Y%5E2-3.5X+-2Y+%2B+3.5+=+0