Suppose we have two circles $S_1$ and $S_2$.
Why is the circle through the point of intersection of the two circles given by:
$S_1 + \lambda(S_2 - S_1) = 0$
and not:
$S_2 + \lambda(S_1) = 0$?
These two questions along with their solutions are the source of my confusion:
Question 1)
Question 2)
It can be clearly seen that in the first question then have used $S_1 + \lambda (S_2) = 0$
And in the next one $S_1 + \lambda (S_1 - S_2) = 0$
So I am unable to understand when to use which form.
On
In general, you want to have a linear combination of the two equations that represents all of the circles (and the line, which you can think of here as a circle with infinite radius) that pass through the two intersection points. If you write this combination as $S_1+\lambda S_2=0$, though, there’s no value of $\lambda$ that will produce the second circle. That’s why one uses the affine (a.k.a. convex) combination $S_1+\lambda(S_2-S_1)$, which I prefer to write as $(1-\lambda)S_1+\lambda S_2$. If you set $\lambda=1$ in that expression, you get $S_2$. Compare this to a common parametric equation for the line through two points $P_1$ and $P_2$: $P_1+\lambda(P_2-P_1)$. The same principle is at work here.
Now, if you happen to know that $S_2$ can’t be a solution to your problem, then you can use $S_1+\lambda S_2$ instead, but if you can’t be sure of that, you need to use the other combination so that you don’t overlook potential solutions. It’s just as easy to use the more inclusive version, so you might as well use that all of the time. It’s also a good exercise to prove to yourself that the two expressions do produce equations for the same sets of circles, other than $S_2$.
Because the coefficients of each circle need to add up to 1 to make it a convex combination -- in the proposed example, they are $1-\lambda$ and $\lambda$ (for $S_1$ and $S_2$, respectively). Having a convex combination ensures that if $p \in S_1 \cap S_2$ then $p$ will also be in the resulting figure.
In your suggestion, they are $\lambda$ and $1$...