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.