The question given is:
Tangents are drawn from two points $(x_1,y_1)$ and $(x_2,y_2)$ to $xy=c^2$. The conic passing through the two points and through the four points of contact will be a circle, then
$(A)\quad x_1x_2=y_1y_2\\(B)\quad x_1y_2=x_2y_2\\(C)\quad x_1y_2+x_2y_1=4c^2\\(D)\quad x_1x_2+y_1y_2=4c^2$
So we have to find a curve which passes through the four points of contacts created due to the two tangents from each of $(x_1,y_1)$ and $(x_2,y_2)$.
I was told that we can consider such a curve as $$(xy_1+yx_1-c^2)(xy_2+yx_2-c^2)+\lambda (xy-c^2)=0$$
But I wanted to know how exactly the second term of the expression came to be. I understand that the first term denotes the curve passing through the chords of contact.
I am aware of how to write equations for families of conics. For eg. To write the equation of a family of circles through 2 circles $S_1$ and $S_2$ we use the notation $$S_1+\lambda S_2$$ because every point that satisfies both of the curves gives a zero.
But what I don't get in my question is that why can we not write the family as for eg. this:
$$(xy_1+yx_1-c^2)(xy-c^2)+\lambda (xy_2+yx_2-c^2)=0$$
Is this simply because this one has 3 degree terms also which would violate the conditions for it to be a circle as we desire in the question? Isn't this also a curve which is satisfied at all points of intersection?
Any help would be appreciated.
The equation (thanks to Jan-Magnus Økland for pointing out a mistake in the first version) $$ (xy_1+yx_1-2c^2)(xy_2+yx_2-2c^2)=0 $$ represents the two lines including both chords of contact (a degenerate conic). The intersection of this conic with the given hyperbola is formed by the four tangency points: hence the equation $$ (xy_1+yx_1-2c^2)(xy_2+yx_2-2c^2)+\lambda (xy-c^2)=0 $$ represents the pencil of conics passing through the tangency points.