A direct relation

Suppose (x1,y1)(x_1 , y_1) ,(x2,y2)(x_2 , y_2) and (x3,y3)(x_3 , y_3) are three given points on a circle. Let its center be denoted by (h,k)(h,k). Then,

h=12(i=1,2,3)(j=2,3,1)(k=3,1,2)(yiyj)(xk2+yk2)(i=1,2,3)(j=2,3,1)xiyj  (i=1,2,3)(j=2,3,1)yixj\boxed{h = \dfrac 12 \cdot \dfrac{\displaystyle \sum_{(i=1,2,3) \\ (j=2,3,1) \\ (k=3,1,2)} \left( y_i - y_j \right) \left( {x_k}^2 + {y_k}^2 \right)} { \displaystyle \sum_{(i=1,2,3) \\ (j=2,3,1)} x_i y_j ~-~ \displaystyle \sum_{(i=1,2,3) \\ (j=2,3,1)} y_i x_j }}

k=12(i=1,2,3)(j=2,3,1)(k=3,1,2)(xixj)(xk2+yk2)(i=1,2,3)(j=2,3,1)yixj  (i=1,2,3)(j=2,3,1)xiyj\boxed{k = \dfrac 12 \cdot \dfrac{\displaystyle \sum_{(i=1,2,3) \\ (j=2,3,1) \\ (k=3,1,2)} \left( x_i - x_j \right) \left( {x_k}^2 + {y_k}^2 \right)} { \displaystyle \sum_{(i=1,2,3) \\ (j=2,3,1)} y_i x_j ~-~ \displaystyle \sum_{(i=1,2,3) \\ (j=2,3,1)} x_i y_j }}

And the equation of circle is:

(xh)2+(yk)2=(hxa)2+(kya)2\boxed{ {(x-h)}^2 + {(y-k)}^2 = {(h-{x_a})}^2 + {(k-{y_a})}^2 } where a=1,2,3a = 1,2,3.

The proof is very simple yet lengthy, so I'll leave it up to the reader to find it out. You can give your proof in the comments. The diagram given below might be helpful.

Note by Tapas Mazumdar
4 years, 9 months ago

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Is there any intuition behind the equations?

Note: It is better to write the indices as xiyi+1 x_i y_{i+1} instead of trying to express what you mean via (i=1,2,3)(j=2,3,1) (i=1,2,3)(j=2,3,1) .

Calvin Lin Staff - 4 years, 9 months ago

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