Analytic proof of equation of circle passing through intersection of two other circles - From Schaum's Guide I'm trying to teach myself calculus and very slowly working through Schaum's guide. This question is asked elsewhere but I'm afraid I need the answer broken down more:
Let $ \mathscr{C_{1}} $ and $ \mathscr C_{2} $ be two intersecting circles determined by the equations $$ x^2 + y^2 + A_{1}x + B_{1}y + C_{1} = 0 $$ and $$ x^2 + y^2 + A_{2}x + B_{2}y + C_{2} = 0. $$ For any number $ k≠−1 $, show that $$ x^2 + y^2 + A_{1}x + B_{1}y + C_{1} + k ( x^2 + y^2 + A_{2}x + B_{2}y + C_{2} ) = 0 $$ is the equation of a circle through the intersection points of $\mathscr{C_{1}}$ and $\mathscr C_{2}$. Show, conversely, that every such circle may be represented by such an equation for a suitable $ k $.


If anyone could give a pretty detailed walk-through on this I would be supremely grateful. I'm supposed to be able to answer/show it using only analytic geometry (no trig or calculus).
Thank you in advance!!!
 A: For the first point, note that if $M_1(x_1,y_1)$ is one of those two points of intersection, then $x_1^2+y_1^2+A_1x_1+B_1y_1+C_1=0$ because $M_1$ is on the first. Likewise $x_1^2+y_1^2+A_2x_1+B_2y_1+C_2=0$ because it's on the other circle.
So for all $\alpha$ and $\beta$ such that at least one of $\alpha$ and $\beta$ is nonzero:
$$\alpha(x_1^2+y_1^2+A_1x_1+B_1y_1+C_1)+\beta(x_1^2+y_1^2+A_2x_1+B_2y_1+C_2)=\alpha\cdot0+\beta\cdot0=0$$
Now let the coordinates vary: $\alpha(x^2+y^2+A_1x+B_1y+C_1)+\beta(x^2+y^2+A_2x+B_2y+C_2)=0$ is the equation of a circle which also contains $M_1$. (divide by $\alpha+\beta$ to get the equation as $x^2+y^2+Ux+Vy+W=0$, which is the equation of a circle.)
Note that if $\alpha=\beta=0$, this is not anymore the equation of a circle, as it simplifies to $0=0$.

Conversely, if you have a circle $\mathscr C$ through these two points of intersection (call them $M_1,M_2$), then pick a third point on the circle, $M_3(x_3,y_3)$, and choose $\alpha$ and $\beta$ such that at least one of $\alpha,\beta$ is nonzero and
$$\alpha(x_3^2+y_3^2+A_1x_3+B_1y_3+C_1)+\beta(x_3^2+y_3^2+A_2x_3+B_2y_3+C_2)=0$$
Then, letting $x$ and $y$ vary, $\alpha(x^2+y^2+A_1x+B_1y+C_1)+\beta(x^2+y^2+A_2x+B_2y+C_2)=0$ is the equation of a circle which goes through the three points $M_1,M_2$ and $M_3$. There is only one such circle, and it's $\mathscr C$. If $\alpha\ne0$, divide by $\alpha$ to get the equation as in your question.
Like in the first part, if $\alpha=\beta=0$, you don't get the equation of a circle, so the additional condition on $\alpha,\beta$ is again necessary.

Note, as pointed out in the comment above, that it's always possible to find such $\alpha,\beta$, but in the form in your question it's not always possible to find such a $k$: for $\mathscr C=\mathscr C_2$ we would need $\alpha=0$ and $\beta\ne1$, which would lead to $k=\infty$).
A: The conditions for a quadratic equation to represent a circle are:


*

*the coefficients of $x^2$ and $y^2$ are equal and nonzero,

*the coefficient of $xy$ is zero.
You easily check that these hold for the three given equations (with a small exception).
Next, you can check that the two intersection points are common to all these circles.
