Equation of Circle with Endpoints of Diameter

Let A\left(  {{x_1},{y_1}} \right) and B\left(  {{x_2},{y_2}} \right) be the end points of the diameter of the circle as shown in the diagram.


equation-circle-diameter

Let P\left( {x,y} \right) be any point of the circle. Connecting the points A and B with the point P and makes an angle {90^ \circ } between them. First we find the slopes of the lines PA and PB as:
Slope of the line PA =  \frac{{y - {y_1}}}{{x - {x_1}}}
Slope of the line PB =  \frac{{y - {y_2}}}{{x - {x_2}}}
Since m\angle APB = {90^ \circ }, so the lines PA and PB are perpendicular to each other, therefore, the product of their slopes is  - 1. i.e.

\begin{gathered} \frac{{y - {y_1}}}{{x - {x_1}}} \times  \frac{{y - {y_2}}}{{x - {x_2}}} =  - 1 \\ \Rightarrow \frac{{\left( {y - {y_1}}  \right)\left( {y - {y_2}} \right)}}{{\left( {x - {x_1}} \right)\left( {x -  {x_2}} \right)}} =  - 1 \\ \Rightarrow \left( {y - {y_1}} \right)\left(  {y - {y_2}} \right) =  - \left( {x -  {x_1}} \right)\left( {x - {x_2}} \right) \\ \end{gathered}


  \Rightarrow \boxed{\left( {x - {x_1}} \right)\left( {x - {x_2}} \right) +  \left( {y - {y_1}} \right)\left( {y - {y_2}} \right) = 0}


This is the equation of the circle through the extremities (ends) of its diameter. In order to find the centre and radius of this circle, we simplify above equation of circle as follows:

\begin{gathered} {x^2} - {x_1}x - {x_2}x + {x_1}{x_2} + {y^2}  - {y_1}y - {y_2}y + {y_1}{y_2} = 0 \\ \Rightarrow {x^2} + {y^2} - \left( {{x_1} +  {x_2}} \right)x - \left( {{y_1} + {y_2}} \right)y + {x_1}{x_2} + {y_1}{y_2} = 0 \\ \end{gathered}


Comparing this equation with the general equation of circle, we have

g  =  - \frac{{{x_1} + {x_2}}}{2},\,\,\,f  =  - \frac{{{y_1} + {y_2}}}{2},\,\,\,c =  {x_1}{x_2} + {y_1}{y_2}


Therefore, the centre of the circle is given by

\left(  { - g, - f} \right) = \left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} +  {y_2}}}{2}} \right)


The radius of the circle is given by

r =  \sqrt {{g^2} + {f^2} - c}  = \sqrt  {{{\left( { - \frac{{{x_1} + {x_2}}}{2}} \right)}^2} + {{\left( { -  \frac{{{y_1} + {y_2}}}{2}} \right)}^2} - \left( {{x_1}{x_2} + {y_1}{y_2}}  \right)}


\begin{gathered} \Rightarrow r = \sqrt {\frac{{{{\left(  {{x_1} + {x_2}} \right)}^2} + {{\left( {{y_1} + {y_2}} \right)}^2} - 4\left(  {{x_1}{x_2} + {y_1}{y_2}} \right)}}{4}} \\ \Rightarrow r = \frac{{\sqrt {{x_1}^2 +  {x_2}^2 + 2{x_1}{x_2} + {y_1}^2 + {y_2}^2 + 2{y_1}{y_2} - 4{x_1}{x_2} -  4{y_1}{y_2}} }}{2} \\ \Rightarrow r = \frac{{\sqrt {{{\left(  {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} }}{2} \\ \end{gathered}

Example: Find the equation of circle through the ends \left(  {5,7} \right) and \left( {1,3}  \right) of its diameter. Also find the centre and radius.
Equation of circles through ends points of its diameter is

\left(  {x - {x_1}} \right)\left( {x - {x_2}} \right) + \left( {y - {y_1}} \right)\left(  {y - {y_2}} \right) = 0


Here from the given points we have values {x_1} = 5,\,\,{x_2} = 1,\,\,{y_1} = 7,\,\,{y_2} = 3
Now substitute these values of the given points in the above equation of circle as

\begin{gathered} \left( {x - 5} \right)\left( {x - 1} \right)  + \left( {y - 7} \right)\left( {y - 3} \right) = 0 \\ \Rightarrow {x^2} - 6x + 5 + {y^2} - 10y +  21 = 0 \\ \Rightarrow {x^2} + {y^2} - 6x - 10y + 26 =  0 \\ \end{gathered}


The centre of the circle is \left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} +  {y_2}}}{2}} \right) = \left( {\frac{{5 + 1}}{2},\frac{{7 + 3}}{2}} \right) =  \left( {3,5} \right)
The radius of circle is

r = \frac{{\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} +  {{\left( {{y_1} - {y_2}} \right)}^2}} }}{2} = \frac{{\sqrt {{{\left( {5 - 1}  \right)}^2} + {{\left( {7 - 3} \right)}^2}} }}{2} = \frac{{4\sqrt 2 }}{2} =  2\sqrt 2

Comments

comments

Posted in: