How does family of curves formula exactly work? I am familiar with the following equations
$$S_1 + \lambda S_2=0$$ for two circles
$$L_1 + \lambda L_2=0$$ for two lines
$$L_1 + \lambda S_1=0$$ for circle and line
But I never really understood how they worked. Now this is a sample questions

A circles touches the parabola $y^2=2x$ at P $(\frac 12, 1)$ and cuts parabola at vertex V. If centre of circle is a Q, find the radius of circle

The formula used here was
$$(x-\frac 12)^2 + (y-1)^2 +\lambda (2x-2y+1)=0$$
Now it’s easy to see that equation is basically hinting at a curve passing through $(1/2, 1)$ and tangent to $(2x-2y+1)$, but how exactly was the form mat determined? How can we tell if this will us gives a circle? Why was the distance formula used in the first part of the equation? Basically I want to know the process of writing such equations .

$C_1:\left(x-\frac 12\right)^2 + (y-1)^2 =0$
is the equation of the circle having centre in $P\left(\frac12,1\right)$ and radius $r=0$ and $C_2:2x-2y+1=0$ is the equation of the line tangent to the parabola $\mathcal{C}:y^2=2x$ at $P$
A linear combination of the two $C_1+\lambda C_2=0$ represents all circles tangent to the parabola $\mathcal{C}$ at $P$ (see image (2) below)
$$\left(x-\frac 12\right)^2 + (y-1)^2+\lambda(2x-2y+1)=0\tag{1}$$
If the circle passes through the origin, the vertex of $\mathcal{C}$, then substitute $(0,0)$ in $(1)$ to get $\lambda=-5/4$.
The circle we are looking for has equation
$$\left(x-\frac{1}{2}\right)^2+(y-1)^2-\frac{5}{4} (2 x-2 y+1)=0$$
simplify
$$2 x^2+2 y^2-7 x+y=0$$
center is $\left(\frac74,-\frac14\right)$ and radius $r=\frac{5\sqrt 2}{4}$

$$...$$


