Family of Circles at Two Points

[Calculuser]

As I was flipping through pages of my analytic geometry book from high school, in circle section I stumbled across the formula of "family of circles intersecting at two points" with two circles ($x^2 + y^2 + D_1 x + E_1 y + F_1 = 0$ , $x^2 + y^2 + D_2 x + E_2 y + F_2 = 0$) known to intersect at two points as follows: $$x^2 + y^2 + D_1 x + E_1 y + F_1 + \lambda (x^2 + y^2 + D_2 x + E_2 y + F_2) = 0, \qquad \lambda \setminus \{-1\} \in \mathcal{R}$$ Though at first glance one derivation seems obvious when we just multiply one of those two equations by $\lambda$ and add them up to get the final form, yet this raises a question to me why the final equation involving $\lambda$ has to pass the points that the two circle equations used in its derivation pass.

In light of these, my question is as to a rigorous derivation of this equation in which involves theorems from linear algebra and vector algebra such as linear independence, orthogonality of two vectors by dot product, etc. How can we approach this with respect to what I mentioned above?

 

[fresh_42]

As I was flipping through pages of my analytic geometry book from high school, in circle section I stumbled across the formula of "family of circles intersecting at two points" with two circles ($x^2 + y^2 + D_1 x + E_1 y + F_1 = 0$ , $x^2 + y^2 + D_2 x + E_2 y + F_2 = 0$) known to intersect at two points as follows: $$x^2 + y^2 + D_1 x + E_1 y + F_1 + \lambda (x^2 + y^2 + D_2 x + E_2 y + F_2) = 0, \qquad \lambda \setminus \{-1\} \in \mathcal{R}$$ Though at first glance one derivation seems obvious when we just multiply one of those two equations by $\lambda$ and add them up to get the final form, yet this raises a question to me why the final equation involving $\lambda$ has to pass the points that the two circle equations used in its derivation pass.

At the intersection, both coordinates fulfill the equations $C_1(x,y)=0$ and $C_2(x,y)=0$. Thus $C_1(x,y)+\lambda C_2(x,y) =0$ holds, too. In the other direction, you need to allow variation of $\lambda$ in order to conclude that the two summands have to vanish.

In light of these, my question is as to a rigorous derivation of this equation in which involves theorems from linear algebra and vector algebra such as linear independence, orthogonality of two vectors by dot product, etc. How can we approach this with respect to what I mentioned above?

It is not quite clear to me what "derivation" means here. What is the purpose of the entire thing?

 

[Calculuser]

It is not quite clear to me what "derivation" means here. What is the purpose of the entire thing?

What I mean by that is a derivation that contains vector and linear algebra properties used as in the derivation of similar problem, "family of lines passing through one point", as shown below even though irrelevant, but I thought useful to share:

Let $d_1:a_1x+b_1y+c_1=0$ and $d_2:a_2x+b_2y+c_2=0$ be two lines passing through $A(x_0,y_0)$; and $P(x,y)$ be a point on a line, namely $d:Ax+By+C=0$, from family of lines. Then $\vec{n_1}=<a_1,b_1>$, $\vec{n_2}=<a_2,b_2>$, $\vec{n}=<A,B>$ become normal vectors to those lines, respectively. $\vec{AP}=<x-x_0,y-y_0>$ is a vector along line $d$. For $d_1$ and $d_2$ are not parallel, they are linearly independent; and $\vec{n}$ vector can be written as a linear combination of them as such: $$\vec{n}=k_1\vec n_1+k_2\vec n_2$$
or $$A=k_1a_1+k_2a_2$$ $$B=k_1b_1+k_2b_2$$
$A(x_0,y_0)$ satisfies the equations, then: $$c_1=-(a_1x_0+b_1y_0)$$ $$c_2=-(a_2x_0+b_2y_0)$$
We also know $A(x_0,y_0)$ satisfies the equation $d$, $$Ax_0+By_0+C=0$$
Plugging $A$ and $B$ relations above into $Ax_0+By_0+C=0$, $$k_1(a_1x_0+b_1y_0)+k_2(a_2x_0+b_2y_0)+C=0$$
or $$-k1c_1-k_2c_2+C=0$$
Then the equation of line $d$ becomes, $$(k_1a_1+k_2a_2)x+(k_1b_1+k_2b_2)y+(k_1c_1+k_2c_2)=0$$
Rearranging terms we get, $$k_1(a_1x+b_1y+c_1)+k_2(a_2x+b_2y+c_2)=0$$
If we define a parameter $\lambda=\frac{k_2}{k_1}, k_1\neq 0$, then the equation becomes, $$(a_1x+b_1y+c_1)+\lambda (a_2x+b_2y+c_2)=0, \lambda \in \mathbb{R} \setminus 0$$
What I am looking for is such rigorous derivation if possible as exemplified above other than:

At the intersection, both coordinates fulfill the equations $C_1(x,y)=0$ and $C_2(x,y)=0$. Thus $C_1(x,y)+\lambda C_2(x,y) =0$ holds, too. In the other direction, you need to allow variation of $\lambda$ in order to conclude that the two summands have to vanish.