14.0 Normal to a circle at a given point

The normal of a circle at any point is a straight line which is perpendicular to the tangent at the point and always passes through the centre of the circle.

The equation of normal at the point $P({x_1},{y_1})$ to the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ $$\frac{{x - {x_1}}}{{{x_1} + g}} = \frac{{y - {y_1}}}{{{y_1} + f}}$$

Proof: Since $P({x_1},{y_1})$ be a point on the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0...(1)$

Let $PT$ be the tangent to the circle at point $P$ and $CP$ be the normal to the circle at $P$, then $CP$ is perpendicular to $PT$. Now equation of tangent to the circle given in equation $(1)$ at $P({x_1},{y_1})$ using point form is $$x{x_1} + y{y_1} + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + c = 0$$ $$x\left( {{x_1} + g} \right) + y\left( {{y_1} + f} \right) + g{x_1} + f{y_1} + c = 0$$

Therefore, slope of tangent $PT = - \frac{{\left( {{x_1} + g} \right)}}{{\left( {{y_1} + f} \right)}}$

And slope of normal $CP = - \frac{1}{{slope\ of\ tangent}} = \frac{{\left( {{y_1} + f} \right)}}{{\left( {{x_1} + g} \right)}}$

Now, equation of normal $CP$ is $$y - {y_1} = \frac{{\left( {{y_1} + f} \right)}}{{\left( {{x_1} + g} \right)}}\left( {x - {x_1}} \right)$$ $$\frac{{x - {x_1}}}{{{x_1} + g}} = \frac{{y - {y_1}}}{{{y_1} + f}}$$

Alternate Method: Using Determinant and valid for all second degree conics.

To find the equation of normal at point $P({x_1},{y_1})$ of second degree conics $$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0...(1)$$

Assume the determinant $\left| {\begin{array}{c}a&h&g \\h&b&f \\g&f&c\end{array}} \right|$ and write first two rows as $a{x_1} + h{y_1} + g$ and $h{x_1} + b{y_1} + f$. Then equation of normal of $(1)$ at $P({x_1},{y_1})$ is $$\frac{{x - {x_1}}}{{a{x_1} + h{y_1} + g}} = \frac{{y - {y_1}}}{{h{x_1} + b{y_1} + f}}$$