1. Based on Circumcenter Vector Derivation
1.1 origin triangle
Assuming \ (\ triangle ABC \) in, \ (A \) coordinates for the \ ((0,0) \) , the coordinates of the remaining points are \ (B (x_1, y_1) , C (x_2, y_2) \ ) , we call \ (\ triangle ABC \) as the origin triangle. For non-triangular origin, can \ (A, B, C \ ) subtracted \ (A \) coordinates, thereby transforming into the origin triangle . In origin triangle \ (\ triangle ABC \) , it is assumed circumcenter \ (P \) coordinates \ ((X, Y) \) .
1.2 Properties of the vector defined circumcenter
\ (\ triangle ABC \) excenter is defined as: circumcenter \ (P \) to \ (\ triangle ABC \) equal to the distance of each vertex, as represented by a vector:
\(|\overrightarrow{PA}|=|\overrightarrow{PB}|=|\overrightarrow{PC}|\)
From the above properties can be obtained:
\ (\ Left \ {\ begin {aligned} | \ overrightarrow {PA} | = | \ overrightarrow {PB} | \ Longrightarrow \ triangle ABP is an isosceles triangle \\ | \ overrightarrow {PA} | = | \ overrightarrow {PC } | \ Longrightarrow \ triangle ACP is an isosceles triangle \\ | \ overrightarrow {PB} | = | \ overrightarrow {PC} | \ Longrightarrow \ triangle BCP is an isosceles triangle \ end {aligned} \ right \).
In general, the isosceles triangle \ (\ triangle ABP \) in, \ (\ overrightarrow the AP {} \) in \ (\ overrightarrow {AB} \ ) projected on to:
\(\cfrac{\overrightarrow{AP}\cdot\overrightarrow{AB}}{|\overrightarrow{AB}|} = \cfrac{|\overrightarrow{AB}|}{2} \Longrightarrow \overrightarrow{AP}\cdot\overrightarrow{AB} = \cfrac{|\overrightarrow{AB}|^2}{2}\)
Therefore, we can get \ (\ triangle ABC \) circumscribed circle vector nature:
\(\left\{\begin{aligned} \overrightarrow{AP}\cdot\overrightarrow{AB} = \cfrac{|\overrightarrow{AB}|^2}{2} \\ \overrightarrow{AP}\cdot\overrightarrow{AC} = \cfrac{|\overrightarrow{AC}|^2}{2} \\ \overrightarrow{BP}\cdot\overrightarrow{BC} = \cfrac{|\overrightarrow{BC}|^2}{2} \end{aligned}\right.\) 式1
1.3 triangle circumcenter coordinate origin derived
The origin triangle \ (\ triangle ABC \) defined can be obtained:
\ (\ left \ {\ the aligned the begin {} \ overrightarrow the AP} = {(X, Y) \\ \ overrightarrow {AB} = (x_1, Y_1) \ \ \ overrightarrow {AC} = ( x_2, y_2) \ end {aligned} \ right. \)
1.2 Properties of the circumcircle derived vector will take the item Substituting equation 2, can be obtained:
\(\left\{\begin{aligned} (x,y)\cdot(x_1,y_1) = \cfrac{|(x_1,y_1)|^2}{2} \\ (x,y)\cdot(x_2,y_2) = \cfrac{|(x_2,y_2)|^2}{2} \end{aligned}\right.\)
\(\Longrightarrow\left\{\begin{aligned} x_1\cdot x+y_1\cdot y = \cfrac{x_1^2+y_1^2}{2} \\ x_2\cdot x+y_2\cdot y = \cfrac{x_2^2+y_2^2}{2} \end{aligned}\right.\)
令\(b_1=\cfrac{x_1^2+y_1^2}{2}, b_2=\cfrac{x_2^2+y_2^2}{2}\)
\(\Longrightarrow\left\{\begin{aligned} x_1\cdot x+y_1\cdot y = b_1 \\ x_2\cdot x+y_2\cdot y = b_2 \end{aligned}\right.\)
Formula can be used to solve linear equations Cramer Rule to give:
\(D=\left|\begin{array}{cccc} x_1 & y_1\\ x_2 & y_2 \end{array}\right| = x_1\cdot y_2 - x_2\cdot y_1\)
\(\Longrightarrow\left\{\begin{aligned} x=\cfrac{\left|\begin{array}{cccc} b_1 & y_1\\ b_2 & y_2 \end{array}\right|}{D} \\ y = \cfrac{\left|\begin{array}{cccc} x_1 & b_1\\ x_2 & b_2 \end{array}\right|}{D} \end{aligned}\right.\)
In the end, we got circumcenter coordinate formula:
\(\Longrightarrow\left\{\begin{aligned} x=\cfrac{b_1\cdot y_2 - b_2\cdot y_1}{D} \\ y = \cfrac{x_1\cdot b_2 - x_2\cdot b_1}{D} \end{aligned}\right.\) 式2
Deriving a generally triangular coordinate circumcenter 1.4
Of a generally triangular \ (\ Triangle A ^ { '} B ^ {'} C ^ { '} \) , with vertices coordinates \ (A (x_0 ^ {' }, y_0 ^ { '}), B (x_1 ^ { '}, Y_1 ^ {'}), C (x_2 ^ { '}, {Y_2 ^'}) \) , the vertex coordinates are subtracted \ (A \) coordinates, to give origin triangle \ (\ triangle the ABC \) and the vertex coordinates \ (A (0,0), B (x_1, Y_1), C (x_2, Y_2) \) .
Application of type 2 , we get a triangle circumcenter coordinate formula
\(\Longrightarrow\left\{\begin{aligned} x = \cfrac{b_1\cdot y_2 - b_2\cdot y_1}{D}+(x_0^{'}-x_0) \\ y = \cfrac{x_1\cdot b_2 - x_2\cdot b_1}{D}+(y_0^{'}-y_0) \end{aligned}\right.\) 式3
2. Using the coordinate origin method to derive the triangle circumcenter
Origin of triangular \ (\ Triangle the ABC \) , the use of the circumscribed circle is defined with:
\(\left\{\begin{aligned} x^2 + y^2 = (x_1 - x)^2 + (y_1 - y)^2\\ x^2 + y^2 = (x_2 - x)^2 + (y_2 - y)^2 \end{aligned}\right.\)
\(\Rightarrow\left\{\begin{aligned} x^2 + y^2 = x_1^2 - 2x_1x + x^2 + y_1^2 - 2y_1y + y^2\\ x^2 + y^2 = x_2^2 - 2x_2x + x^2 + y_2^2 - 2y_2y + y^2 \end{aligned}\right.\)
\(\Rightarrow\left\{\begin{aligned} 0 = x_1^2 - 2x_1x + y_1^2 - 2y_1y\\ 0 = x_2^2 - 2x_2x + y_2^2 - 2y_2y \end{aligned}\right.\)
\(\Rightarrow\left\{\begin{aligned} 2x_1x + 2y_1y = x_1^2 + y_1^2\\ 2x_2x + 2y_2y = x_2^2 + y_2^2 \end{aligned}\right.\)
\(\Rightarrow\left\{\begin{aligned} x_1\cdot x + y_1\cdot y = \cfrac{x_1^2 + y_1^2}{2}\\ x_2\cdot x + y_2\cdot y = \cfrac{x_2^2 + y_2^2}{2} \end{aligned}\right.\)
Here, the same results obtained with the deduced Vector \ (♡ \) .