Tangent at a Point on the Circle

Equation of the tangent to x² + y² + 2gx + 2fy + c = 0 at the point (x₁, y₁) lying on it. Obtained by the T = 0 substitution rule.

Let the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ with centre $C(-g, -f)$, and let $P(x_1, y_1)$ be a point on it.

The radius $CP$ has slope:

$$m_{CP} = \frac{y_1 - (-f)}{x_1 - (-g)} = \frac{y_1 + f}{x_1 + g}$$

The tangent at $P$ is perpendicular to $CP$, so its slope is:

$$m_T = -\frac{x_1 + g}{y_1 + f}$$

Equation of the tangent through $P(x_1, y_1)$:

$$y - y_1 = -\frac{x_1 + g}{y_1 + f}(x - x_1)$$

Multiplying through by $(y_1 + f)$:

$$(y - y_1)(y_1 + f) = -(x_1 + g)(x - x_1)$$

Expanding and collecting:

$$xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$$

The last step uses the fact that $(x_1, y_1)$ lies on the circle: $x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c = 0$.

The substitution rule (T = 0): To write the tangent at $(x_1, y_1)$, replace $x^2 \to xx_1$, $y^2 \to yy_1$, $2x \to x + x_1$, $2y \to y + y_1$ in the circle equation. This pattern holds for all conics.