Q.Find the equation of the circle described on the common chord of the circles $x^2+y^2-4x-2y+4=0$ and $x^2+y^2-2x-4y+4=0$ as diameter.

Given two circles $S\equiv x^2+y^2+2gx+2fy+c=0$ and $S'\equiv x^2+y^2+2g'x+2f'y+c'=0$, the equation

represents, for each value of the parameter $\lambda$, a circle (the coefficients of $x^2$ and $y^2$ are both $1+\lambda$, equal to each other, so after dividing through by $1+\lambda$ the equation is in standard circle form) that passes through every point common to $S=0$ and $S'=0$ -- because any such common point satisfies $S=0$ and $S'=0$ simultaneously, hence also $S+\lambda S'=0$ for every $\lambda$. Choosing $\lambda$ so that this family member also passes through some further prescribed point (or satisfies some other condition, such as its centre lying on a given line) singles out one particular circle from the family.

The same idea works when one of the two curves is a straight line $L\equiv lx+my+n=0$ rather than a circle: the family

$S+kL=0$ …