The Midpoint of the Hypotenuse is the Circumcentre of the Right Triangle

The midpoint of the hypotenuse of a right triangle is the circum-centre of the triangle.


hypotenuse-circumcentre-right-triangle

Consider the equation of the circle in general form is given by the equation

{x^2}  + {y^2} + 2gx + 2fy + c = 0\,\,\,\,{\text{ -   -  - }}\left( {\text{i}} \right)


Let A\left( {a,0} \right), B\left( {b,0} \right) and C\left( {b,c} \right) are any three point on the given circle.
For the point A\left(  {a,0} \right), since the point A is on the circle then the equation of circle becomes

\begin{gathered} {\left( a \right)^2} + {\left( 0 \right)^2} +  2g\left( a \right) + 2f\left( 0 \right) + c = 0 \\ {a^2} + 2ag + c = 0\,\,\,\,{\text{ -  -  -  }}\left( {{\text{ii}}} \right) \\ \end{gathered}


For the point B\left(  {b,0} \right), since the point B is on the circle then the equation of circle becomes

\begin{gathered} {\left( b \right)^2} + {\left( 0 \right)^2} +  2g\left( b \right) + 2f\left( 0 \right) + c = 0 \\ {b^2} + 2bg + c = 0\,\,\,\,{\text{ -  -  -  }}\left( {{\text{iii}}} \right) \\ \end{gathered}


For the point C\left(  {b,c} \right), since the point C is on the circle then the equation of circle becomes

\begin{gathered} {\left( b \right)^2} + {\left( c \right)^2} +  2g\left( b \right) + 2f\left( c \right) + c = 0 \\ {b^2} + {c^2} + 2bg + 2cf + c =  0\,\,\,\,{\text{ -  -  - }}\left( {{\text{iv}}} \right) \\ \end{gathered}


Now solving equation (iv) and equation (iii), we get the value

\begin{gathered} 2cf =   - {c^2} \\ f =  -  \frac{{{c^2}}}{{2c}} \Rightarrow f =  -  \frac{c}{2} \Rightarrow  - f =  \frac{c}{2} \\ \end{gathered}


Solving equation (ii) and equation (iii), we get the value

\begin{gathered} \left( {{a^2} - {b^2}} \right) + \left( {2a -  2b} \right)g = 0 \\ \Rightarrow 2\left( {a - b} \right)g =  - \left( {{a^2} - {b^2}} \right) \\ \Rightarrow g = \frac{{ - \left( {a - b}  \right)\left( {a + b} \right)}}{{2\left( {a - b} \right)}} \\ \Rightarrow   - g = \frac{{a + b}}{2} \\ \end{gathered}


Centre of the circle is given by

\left(  { - g, - f} \right) = \left( {\frac{{a + b}}{2},\frac{c}{2}} \right)


Now midpoint of the hypogenous is given as

\begin{gathered} AC = \left( {\frac{{{x_1} +  {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right) \\ AC = \left( {\frac{{a + b}}{2},\frac{{0 +  c}}{2}} \right) = \left( {\frac{{a + b}}{2},\frac{c}{2}} \right) \\ \end{gathered}


Thus, midpoint of the hypotenuse is equal to centre of the circle.

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