# Normal Line of a Circle Passes through the Origin

The normal lines of a circle passes through the center of the circle.

Consider the equation of the circle
${x^2} + {y^2} = {r^2}\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Since the circle passes through the point $A\left( {{x_1},{y_1}} \right)$, the equation of the circle becomes
${x_1}^2 + {y_1}^2 = {r^2}\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

Now differentiating the equation of a circle (i) with respect to $x$, we have
$\begin{gathered} \frac{d}{{dx}}{x^2} + \frac{d}{{dx}}{y^2} = \frac{d}{{dx}}{r^2} \\ \Rightarrow 2x + 2y\frac{{dy}}{{dx}} = 0 \\ \Rightarrow x + y\frac{{dy}}{{dx}} = 0 \Rightarrow y\frac{{dy}}{{dx}} = – x \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{x}{y} \\ \end{gathered}$

If ${m_1}$ is the slope of the tangent line at point $A\left( {{x_1},{y_1}} \right)$, then
${m_1} = {\frac{{dy}}{{dx}}_{\left( A \right)}} = – \frac{{{x_1}}}{{{y_1}}}$

The slope of the normal line at the point $A\left( {{x_1},{y_1}} \right)$ is $m = – \frac{1}{{{m_1}}} = \frac{{{y_1}}}{{{x_1}}}$

Now the equation of the normal at the point $A\left( {{x_1},{y_1}} \right)$ using she tlope point is
$\begin{gathered} y – {y_1} = m\left( {x – {x_1}} \right) \\ \Rightarrow y – {y_1} = \frac{{{y_1}}}{{{x_1}}}\left( {x – {x_1}} \right) \\ \Rightarrow {x_1}y – {x_1}{y_1} = {y_1}x – {x_1}{y_1} \\ \Rightarrow {x_1}y = {y_1}x\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right) \\ \end{gathered}$

Now putting the values $x = 0,\,\,y = 0$ in equation (iii), we get the result
${x_1}\left( 0 \right) = {y_1}\left( 0 \right) \Rightarrow 0 = 0$

This shows that the normal line passes through the center $\left( {0,0} \right)$ of the circle.