Family of Circles Through Intersection of Two Circles

For λ ≠ −1, this represents a circle passing through the two intersection points of S₁ = 0 and S₂ = 0. When λ = −1 it degenerates to the radical axis.

Let $S_1 = 0$ and $S_2 = 0$ be two circles intersecting at points $A$ and $B$.

Consider $S_1 + \lambda S_2 = 0$ for a real parameter $\lambda \neq -1$:

$$(x^2 + y^2 + 2g_1x + 2f_1y + c_1) + \lambda(x^2 + y^2 + 2g_2x + 2f_2y + c_2) = 0$$ $$(1+\lambda)x^2 + (1+\lambda)y^2 + \cdots = 0$$

For $\lambda \neq -1$, the coefficients of $x^2$ and $y^2$ are equal and there is no $xy$ term — this represents a circle.

It passes through $A$ and $B$: Since $S_1(A) = 0$ and $S_2(A) = 0$, we have $S_1(A) + \lambda S_2(A) = 0$ for any $\lambda$. Same for $B$. So every member of the family passes through both intersection points.

$\lambda = -1$: The equation degenerates to $S_1 - S_2 = 0$, which is linear — this is the radical axis (the common chord $AB$ extended).

Choosing $\lambda$: One free parameter means one additional condition can be imposed (e.g., passes through a specific third point, has a given radius, or has its centre on a given line). Substitute the condition to solve for $\lambda$.

When the circles do not intersect: The family $S_1 + \lambda S_2 = 0$ still makes algebraic sense but has no real common points. The circles are part of a coaxial system, and the radical axis still exists as a real line.