Length of the Diameter of a Circle

The length of the diameter of the circle {x^2} + {y^2} = {r^2} is equal to 2r.

Consider the equation of a circle is given by

{x^2} + {y^2} = {r^2}\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right)


length-of-diameter-circle

Let A\left( {{x_1},{y_1}} \right) be the point of the circle. By putting this point in a circle, then

{x_1}^2 + {y_1}^2 = {r^2}\,\,\,\,{\text{ - - - }}\left( {{\text{ii}}} \right)

Since the origin O\left( {0,0} \right) is the center of the given circle (i), the line through O and A across the circle is its diameter. The equation of the line through O and A using two points from the line is

\begin{gathered} \frac{{y - 0}}{{{y_1} - 0}} = \frac{{x - 0}}{{{x_1} - 0}} \\ \Rightarrow y = \frac{{{y_1}x}}{{{x_1}}}\,\,\,{\text{ - - - }}\left( {{\text{iii}}} \right) \\ \end{gathered}

Putting the value of y from equation (iii) in equation (i), we have

\begin{gathered} {x^2} + \left( {\frac{{{y_1}x}}{{{x_1}}}} \right) = {r^2} \\ \Rightarrow {x^2} + \frac{{{y_1}^2{x^2}}}{{{x_1}^2}} = {r^2} \\ \Rightarrow \frac{{{x_1}^2{x^2} + {y_1}^2{x^2}}}{{{x_1}^2}} = {r^2} \\ \Rightarrow \frac{{\left( {{x_1}^2 + {y_1}^2} \right){x^2}}}{{{x_1}^2}} = {r^2} \\ \Rightarrow \frac{{\left( {{r^2}} \right){x^2}}}{{{x_1}^2}} = {r^2}\,\,\,\,\,\,\,\because {x_1}^2 + {y_1}^2 = {r^2} \\ \Rightarrow \frac{{{x^2}}}{{{x_1}^2}} = 1 \Rightarrow {x^2} = {x_1}^2 \\ \Rightarrow x = \pm {x_1} \\ \end{gathered}

Putting x = - {x_1} in the above equation (iii), we have y = \frac{{{y_1}\left( { - {x_1}} \right)}}{{{x_1}}} = - {y_1}. This shows that the other end of the diameter is B\left( { - {x_1},{y_1}} \right), as shown in the given diagram. Now the length of the diameter is

\begin{gathered} \left| {AB} \right| = \sqrt {{{\left( {{x_1} - \left( { - {x_1}} \right)} \right)}^2} + {{\left( {{y_1} - \left( { - {y_1}} \right)} \right)}^2}} = \sqrt {{{\left( {{x_1} + {x_1}} \right)}^2} + {{\left( {{y_1} + {y_1}} \right)}^2}} \\ \,\,\,\,\,\,\,\,\,\, = \sqrt {{{\left( {2{x_1}} \right)}^2} + {{\left( {2{y_1}} \right)}^2}} = \sqrt {4{x_1}^2 + 4{y_1}^2} = \sqrt {4\left( {{x_1}^2 + {y_1}^2} \right)} \\ \,\,\,\,\,\,\,\,\,\, = \sqrt {4{r^2}} = 2r \\ \end{gathered}

This is the required result.