# 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

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

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.