Normal Line of Circle Passes through Origin

The normal lines of a circle pass through the centre 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) so equation of circle becomes

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


Now differentiating equation of 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 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}}}


Slope of normal line at the point A\left( {{x_1},{y_1}} \right) is m =  -  \frac{1}{{{m_1}}} = \frac{{{y_1}}}{{{x_1}}}
Now equation of normal at the point A\left( {{x_1},{y_1}} \right) using slope 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 centre \left( {0,0} \right) of the circle.

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