a.maths@

策略
因為三點定一圓﹐只要用頭三點作了一個圓﹐而第四點又在其上﹐便可證得四點共圓
Let the equation of the circle by (2,2) , (1,1) ,(4,-2) is
x^2+y^2+Dx+Ey+F=0
sub (2,2) , (1,1) ,(4,-2)
2D+2E+F+8=0...(1)
D+E+F+2=0...(2)
4D-2E+F+20=0...(3)
(1)-(2)*2
-F+4=0
F=4
(3)-(1)*2
-6E-F+4=0
That is -6E=0
So
D+E+6=0...(4)
-6E=0...(5)
Solve it E=0
D=-6
the equation of the circle by (2,2) , (1,1) ,(4,-2) is
x^2+y^2-6x+4=0
sub (5,1) into x^2+y^2-6x+4
5^2+1-6*5+4
=25+1-30+4
=0
So (5,1) is on the circle
We prove that the four point(2,2) , (1,1) ,(4,-2) and (5,1) are concyclic.

(2 - 1) / (2 - 5) = -1/3

(2 - 1) / (2 - 1) = 1

(1 + 2) / (5 - 4) = 3

(1 + 2) / (1 - 4) = -1

there are two 90 degrees angles