Normal at a Point on the Circle

Normal to x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁). Every normal to a circle passes through its centre (−g, −f).

The normal to a curve at a point is the line through that point perpendicular to the tangent there.

For a circle, the tangent at any point is perpendicular to the radius at that point. The normal is therefore parallel to the radius — in fact, the normal is the radius extended.

Equation: For the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ with centre $C(-g, -f)$, the normal at $P(x_1, y_1)$ passes through both $P$ and $C$.

Slope of $CP$:

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

Line through $P(x_1, y_1)$ with this slope:

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

This can be rewritten as:

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

Key fact: All normals of a circle are concurrent — they all pass through the centre. This property distinguishes circles from all other conics, where normals are concurrent only in degenerate cases.