A Seemingly Short and Sweet but Intriguing Christmas Special Circle Problem. (Revised)

Taking the circle equation and completing squares:

$x^{2}+y^{2}-2x-6y-10=0$

$(x^{2}-2x)+(y^{2}-6y)=10$

$(x^{2}-2x)+1+(y^{2}-6y)+9=10+1+9$

$(x-1)^{2}+(y-3)^{2}=20$

Thus, the centre of the circumference is $A=(1,3)$ and its radius is $r=\sqrt{20}$ .

As we are looking for the equations of two parallel lines ( $g$ , in green and $h$ , in blue) to a third line ( $f=y=2x$ , in red), their equations are translations of $f$ through any horizontal axis (in yellow). So, its equations are $2x+a$ and $2x+b$ , for any $a$ and $b$ that satisfies the equations.

Since $\bar{AB}\perp g$ , $\bar{AC}\perp h$ and $g\parallel h$ , $B$ and $C$ are collinear, $B \hat{A} D=C \hat{A} E$ and $C \hat{A} D=B \hat{A} E$ (opposite vertex property).

Now, we can write the coordinates of the tangency points as follows:

$B=(1+r.cos \ B \hat{A} D,\ 3+r.sen \ B \hat{A} D )$

$C=(1+r.cos \ C \hat{A} D,\ 3+r.sen \ C \hat{A} D )$

But $cos \ B \hat{A} D=-cos \ C \hat{A} D$ and $sen \ B \hat{A} D=-sen\ C \hat{A} D$ :

$B=(1+r.cos \ B \hat{A} D,\ 3-r.sen \ C \hat{A} D )=(1+p,\ 3-q)$

$C=(1-r.cos \ B \hat{A} D,\ 3+r.sen \ C \hat{A} D )=(1-p, \ 3+q)$

On the tangency points, the lines' equations are equal to the points above. So, we can write them as:

$2.{x_1}+a={y_1} \rightarrow 2.(1+p)+a=3-q$

$2.{x_2}+b={y_2} \rightarrow 2.(1-p)+b=3+q$ .

Adding the two equations up:

$2.(1+p)+a+2.(1-p)+b=3-q+3+q \therefore \boxed{a+b=2}$ .