# Equations of Tangent and Normal to the Circle

The equations of tangent and normal to the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ at the point $\left( {{x_1},{y_1}} \right)$ are defined by $x{x_1} + y{y_1} + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + c = 0$ and $\left( {y – {y_1}} \right)\left( {{x_1} + g} \right) = \left( {x – {x_1}} \right)\left( {{y_1} + f} \right)$ respectively.

Equation of Tangent to the Circle:
The given equation of a circle is

${x^2} + {y^2} + 2gx + 2fy + c = 0\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Since the given point lies on the circle, it must satisfy (i). We have

${x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c = 0\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

Differentiating both sides of (i) of circle with respect to $x$, we have

$\begin{gathered} 2x + 2y\frac{{dy}}{{dx}} + 2g + 2f\frac{{dy}}{{dx}} + 0 = 0 \\ \Rightarrow x + y\frac{{dy}}{{dx}} + g + f\frac{{dy}}{{dx}} = 0 \\ \Rightarrow \left( {x + y} \right)\frac{{dy}}{{dx}} = – x – g \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{{x + g}}{{y + f}} \\ \end{gathered}$

If $m$ is the slope of the tangent at $\left( {{x_1},{y_1}} \right)$, then

$m = {\frac{{dy}}{{dx}}_{\left( {{x_1},{y_1}} \right)}} = – \frac{{{x_1} + g}}{{{y_1} + f}}$

The equation of the tangent to the circle (i) at the point $\left( {{x_1},{y_1}} \right)$ is

$\begin{gathered} y – {y_1} = m\left( {x – {x_1}} \right) \\ \Rightarrow y – {y_1} = – \frac{{{x_1} + g}}{{{y_1} + f}}\left( {x – {x_1}} \right) \\ \Rightarrow \left( {y – {y_1}} \right)\left( {{y_1} + f} \right) = – \left( {x – {x_1}} \right)\left( {{x_1} + g} \right) \\ \Rightarrow y\left( {{y_1} + f} \right) – {y_1}\left( {{y_1} + f} \right) = – x\left( {{x_1} + g} \right) + {x_1}\left( {{x_1} + g} \right) \\ \Rightarrow y{y_1} + fy – {y_1}^2 – f{y_1} = – x{x_1} – gx + {x_1}^2 + g{x_1} \\ \Rightarrow x{x_1} + y{y_1} + gx + fy = {x_1}^2 + {y_1}^2 + g{x_1} + f{y_1} \\ \end{gathered}$

Adding $g{x_1} + f{y_1} + c$ to both sides, we have

$\begin{gathered} x{x_1} + y{y_1} + gx + fy + g{x_1} + f{y_1} + c = {x_1}^2 + {y_1}^2 + g{x_1} + f{y_1} + g{x_1} + f{y_1} + c \\ \Rightarrow x{x_1} + y{y_1} + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + c = {x_1}^2 + {y_1}^2 + 2g{x_1} + 2g{y_1} + c \\ \Rightarrow x{x_1} + y{y_1} + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + c = 0 \\ \end{gathered}$

This is the equation of the tangent to the circle (i) at point $\left( {{x_1},{y_1}} \right)$.

Equation of Normal to the Circle:

The slope of normal at point $\left( {{x_1},{y_1}} \right)$ is

$– \frac{1}{m} = \frac{{{y_1} + f}}{{{x_1} + g}}$

The equation of normal at $\left( {{x_1},{y_1}} \right)$ is

$\begin{gathered} y – {y_1} = \frac{{{y_1} + f}}{{{x_1} + g}}\left( {x – {x_1}} \right) \\ \Rightarrow \left( {y – {y_1}} \right)\left( {{x_1} + g} \right) = \left( {x – {x_1}} \right)\left( {{y_1} + f} \right) \\ \end{gathered}$

This is the equation of normal to the circle (i) at point $\left( {{x_1},{y_1}} \right)$.