Chord with a Given Midpoint

Equation of the chord of circle S = 0 whose midpoint is (x₁, y₁). Explicitly: xx₁ + yy₁ + g(x+x₁) + f(y+y₁) + c = x₁² + y₁² + 2gx₁ + 2fy₁ + c.

Let $M(x_1, y_1)$ be the midpoint of a chord of circle $x^2 + y^2 + 2gx + 2fy + c = 0$ with centre $C(-g, -f)$.

The perpendicular from the centre bisects the chord, so $CM \perp$ chord.

Slope of $CM$:

$$m_{CM} = \frac{y_1 + f}{x_1 + g}$$

Slope of the chord (perpendicular to $CM$):

$$m_{\text{chord}} = -\frac{x_1 + g}{y_1 + f}$$

Equation of the chord through $M(x_1, y_1)$:

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

Cross-multiplying and expanding:

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

Rearranging using the fact that $(x_1, y_1)$ is the midpoint (not necessarily on the circle):

$$xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$$

In shorthand: $T = S_1$.

This compact identity — $T = S_1$ — is the standard result for the chord with a given midpoint.