Intersection of a Line with a Circle: the Family S + λL = 0

The idea. Suppose a line $L\equiv lx+my+n=0$ cuts a circle $S\equiv x^2+y^2+2gx+2fy+c=0$ at two points $A$ and $B$. There are infinitely many other circles that also pass through the same two points $A,B$ — picture rotating a circle of varying size through the fixed pair of points. Remarkably, every one of them, plus the line itself as a limiting/degenerate case, is captured by one formula:

Why this works: at both $A$ and $B$, $S=0$ and $L=0$ simultaneously, so $S+\lambda L = 0+\lambda\cdot 0=0$ automatically holds at both points, for any real number $\lambda$. And $S+\lambda L$ expands to $x^2+y^2+(2g+\lambda l)x+(2f+\lambda m)y+(c+\lambda n)=0$, which is still a genuine circle equation (coefficient of $x^2$ and $y^2$ both $1$, no $xy$ term) for every $\lambda$. So as $\lambda$ ranges over all real numbers, $S+\lambda L=0$ sweeps out the entire family of circles through $A$ and $B$.

This is extremely useful because you never need to actually solve for $A$ and $B$'s coordinates. Any extra condition on the circle you want (passes through a third point, has its centre on a given line, has a given radius, is orthogonal to another circle, ...) becomes one equation in the single unknown $\lambda$.

Application — the circle with a chord as diameter. A very common question: "line $L$ cuts circle $S$ at $A,B$; find the circle having $\overline{AB}$ as diameter." The circle we want is a member of the family $S+\lambda L=0$ — so all we need is the right $\lambda$. The circle on $AB$ as diameter has its centre at the midpoint of $AB$, and that midpoint clearly lies on the line $L$ itself (since $A,B\in L$). So the extra condition is simply: the centre of $S+\lambda L=0$ must lie on the line $L$. That's one linear equation in $\lambda$ — solve it, substitute back, done.

Worked example. The line $L\equiv x+y-5=0$ cuts the circle $S\equiv x^2+y^2-25=0$ at points $A$ and $B$. Find the equation of the circle having $\overline{AB}$ as diameter.

Solution. The family through $A,B$ is

whose centre is $\left(-\dfrac{\lambda}{2},-\dfrac{\lambda}{2}\right)$. For the diameter-circle, this centre must satisfy $L=0$, i.e. lie on $x+y=5$:

Substituting $\lambda=-5$:

$x^2+y^2-5x-5y-(25-25)=0 \ \Longrightarrow\ \boxed{x^2+y^2-5x-5y=0}$ …