Question

Determine the equation of family of circles passing through two given points $A(x_1,y_1)$ and $B(x_2,y_2)$ . If fixed circle $S_1$ cuts each member of the family in various chords then prove that all these chords pass through a fixed point.

Answer 1

We can determine the equation of a circle passing through three points as its equation involves three unknowns g, f and c. However, if only two points A and B are given, we can find the family of circles given by
S + $\lambda$L = 0 .(1)
where S is circle on AB as diameter and L is the line AB.
S${}_1$ = 0 ..(2)
is another fixed circle.
Common chords of (1) and (2) are given by
Rule: S${}_1{S}_2$ = 0 i.e., (S + $\lambda L)S_1$ = 0
or (S S${}_1)+\lambda$ L = 0 or U + $\lambda$L = 0,
where U is common chord of S and $S_1$ which is fixed. But U + $\lambda$L = 0 represents a family of straight lines (i.e., common chords) which pass through the intersection of fixed lines U = 0 and L = 0 and hence a fixed point.