Equation of a Chord to a Circle If equation of a origin centered $\mathcal circle$  with radius $3$ is $x^2+y^2=9$, what is the $\pmb {equation}$ $\pmb{of ~ the ~chord}$ to this circle which is divided internally by the point $P \equiv (1,2)$ in the ratio $\color{blue} 1: \color{red} 2$?
I've been trying to solve this problem as it appears in my textbook's Circle chapter.
My Attempts

I used section formula and found out if point $A(x_1, y_1)$ and $B(x_2, y_2)$ is the endpoints of the chord $\mathbf {AB}$ (which is also the intersection of the chord and the circle) then using section formula for internal division yields $$\frac{1 \cdot x_2 +2 \cdot x_1}{1+2}=1 \implies 2x_1+x_2=3 \implies x_1+x_2=3- x_1 \qquad (1) $$ and $$ \frac{1 \cdot y_2 +2 \cdot y_1}{1+2} =2 \implies 2y_1+y_2=6  \implies y_1+y_2=6- y_1 \qquad (2) $$ as the point $P(1,2)$ divides the segment (chord) $\mathbf {AB}$ in the ratio $\color{blue} 1: \color{red} 2$ . In the next step, I let the equation of the chord to be $y-2=\frac{y_1-y_2}{x_1 -x_2}(x-1)$ as it passes through the point $P(1,2)$ and $A(x_1, y_1)$, $B(x_2, y_2)$. Then if the center of the circle (which is also the origin) is $O(0, 0)$ the distance $OA$ and $OB$ should be the same as they are the radius of the circle. So, $$ OA^2 = OB^2 \implies x_1^{2} +y_1^{2} = x_2^{2} +y_2^{2} \\ \implies x_1^{2} - x_2^{2} = -(y_1^{2} -y_2^{2}) \\ \implies (x_1 +x_2)(x_1 -x_2)=-(y_1-y_2)(y_1+y_2) \\ \implies \frac{y_1-y_2}{x_1 -x_2} = - \frac{x_1 +x_2}{y_1+y_2} $$
Now, if I plug in the values from $(1)$ and $(2)$ then it becomes- $$\frac{y_1-y_2}{x_1 -x_2} = - \frac{3- x_1}{6- y_1}$$
Now, putting it in the equation of the chord $y-2=\frac{y_1-y_2}{x_1 -x_2}(x-1)$ gives $$y-2=-\frac{3- x_1}{6- y_1} (x-1)$$ Sadly I have $x_1$ and $y_1$ left in my equation.

Was my stride taking me some places or should I change my viewpoint to this particular problem? To be honest, I really can not proceed any further. Please provide
some help.

 A: From 1) and 2) you got
$$
\left\{ \matrix{
  2x_{\,1}  + x_{\,2}  = 3 \hfill \cr 
  2y_{\,1}  + y_{\,2}  = 6 \hfill \cr}  \right.
$$
Now add the condition that A,B are on the circle
$$
\eqalign{
  & \left\{ \matrix{
  2x_{\,1}  + x_{\,2}  = 3 \hfill \cr 
  2y_{\,1}  + y_{\,2}  = 6 \hfill \cr 
  x_{\,1} ^{\,2}  + y_{\,1} ^{\,2}  = 9 \hfill \cr 
  x_{\,2} ^{\,2}  + y_{\,2} ^{\,2}  = 9 \hfill \cr}  \right.\quad \left\{ \matrix{
  2x_{\,1}  + x_{\,2}  = 3 \hfill \cr 
  2y_{\,1}  + y_{\,2}  = 6 \hfill \cr 
  x_{\,1} ^{\,2}  + y_{\,1} ^{\,2}  = 9 \hfill \cr 
  \left( {3 - 2x_{\,1} } \right)^{\,2}  + \left( {6 - 2y_{\,1} } \right)^{\,2}  = 9 \hfill \cr}  \right.  \cr 
  & \left\{ \matrix{
  2x_{\,1}  + x_{\,2}  = 3 \hfill \cr 
  2y_{\,1}  + y_{\,2}  = 6 \hfill \cr 
  x_{\,1} ^{\,2}  + y_{\,1} ^{\,2}  = 9 \hfill \cr 
  9 + 4x_{\,1} ^{\,2}  - 12x_{\,1}  + 36 + 4y_{\,1} ^{\,2}  - 24y_{\,1}  = 9 \hfill \cr}  \right.  \cr 
  & \left\{ \matrix{
  2x_{\,1}  + x_{\,2}  = 3 \hfill \cr 
  2y_{\,1}  + y_{\,2}  = 6 \hfill \cr 
  x_{\,1} ^{\,2}  + y_{\,1} ^{\,2}  = 9 \hfill \cr 
  36 - 12x_{\,1}  - 24y_{\,1}  + 36 = 0 \hfill \cr}  \right.  \cr 
  & \left\{ \matrix{
  2x_{\,1}  + x_{\,2}  = 3 \hfill \cr 
  2y_{\,1}  + y_{\,2}  = 6 \hfill \cr 
  x_{\,1} ^{\,2}  + y_{\,1} ^{\,2}  = 9 \hfill \cr 
  x_{\,1}  + 2y_{\,1}  = 6 \hfill \cr}  \right.\quad \quad \left\{ \matrix{
  2x_{\,1}  + x_{\,2}  = 3 \hfill \cr 
  2y_{\,1}  + y_{\,2}  = 6 \hfill \cr 
  2y_{\,1}  + x_{\,1}  = 6 \hfill \cr 
  x_{\,1} ^{\,2}  + y_{\,1} ^{\,2}  = 9 \hfill \cr}  \right.  \cr 
  & \left\{ \matrix{
  x_{\,2}  = 3 - 2x_{\,1}  \hfill \cr 
  y_{\,1}  = 3 - {{x_{\,1} } \over 2} \hfill \cr 
  y_{\,2}  = x_{\,1}  \hfill \cr 
  x_{\,1} ^{\,2}  + y_{\,1} ^{\,2}  = 9 \hfill \cr}  \right.\quad \left\{ \matrix{
  x_{\,2}  = 3 - 2x_{\,1}  \hfill \cr 
  y_{\,1}  = 3 - {{x_{\,1} } \over 2} \hfill \cr 
  y_{\,2}  = x_{\,1}  \hfill \cr 
  x_{\,1} ^{\,2}  + 9 + {{x_{\,1} ^{\,2} } \over 4} - 3x_{\,1}  = 9 \hfill \cr}  \right.  \cr 
  & \left\{ \matrix{
  x_{\,2}  = 3 - 2x_{\,1}  \hfill \cr 
  y_{\,1}  = 3 - {{x_{\,1} } \over 2} \hfill \cr 
  y_{\,2}  = x_{\,1}  \hfill \cr 
  \left( {{5 \over 4}x_{\,1}  - 3} \right)x_{\,1}  = 0 \hfill \cr}  \right. \cr} 
$$
I suppose you can conclude from here
A: Hm, the problem is that you equated the two evaluated circle equations like so : $$x_1^2 +y_1^2 = x_2^2 + y_2^2$$
This made you lose out a bit of information. To recover, it reintroduce this condition:
$$ x_1^2 + y_1^2 =1 \tag{1}$$
$$x_2^2 + y_2^2=1 \tag{2}$$
We also have,
$$ 2x_1 + x_2=3 \tag{3}$$
$$ 2y_1 + y_2=6 \tag{4}$$
Our goal is to get an expression for $x_1$ and $y_1$, the best way to do it is to isolate for $x_2 $ and $y_2$ in (3) and (4)  and plug those back into $(1)$ and $(2)$:
$$ x_2 = 3-2x_1\tag{5}$$
and,
$$ y_2=3-2y_1 \tag{6}$$
Plugging these into (2):
$$ (3-2x_1)^2 + (3-2y_1)^2 = 1$$
$$ 9+ 4x_1^2 -12x_1 + 9 + 4y_1^2 -12y_1 = 1$$
Or,
$$ 17 + 4(x_1^2 + y_1^2) -12(x_1 + y_1)=0$$
Applying equation $(1)$:
$$ 21 - 12 (x_1 + y_1)=0$$
$$ \frac{7}{4} = x_1 + y_1 \tag{7}$$
Now, equation (7) and (1) can be taken together to solve for $x_1$ and $y_1$
A: Calling
$$
\cases{
A = (x_a,y_a)\\
B = (x_b,y_b)\\
C = (0,0)\\
P = (1,2)\\
r = 3\\
\lambda = 2\\
}
$$
we have the relations
$$
\cases{
\lambda(P-B) = A-P\\
\|A-C\|^2= r^2\\
\|B-C\|^2= r^2\\
}
$$
or
$$
\cases{
2(1-x_b)+1-x_a=0\\ 2(2-y_b)+2-y_a=0\\ x_a^2+y_a^2=9\\ x_b^2+y_b^2=9
}
$$
Four equations and four unknowns. Follows a plot showing (in red) the two solutions

