Finding complex number for intersection of a line and a circle? How to find complex number for intersection of a line and a circle?
In particular, if $M=$ midpoint $BC$, then I want $AM ∩ (ABC)$ in complex numbers.
Let me describe why I needed this,
I was doing a problem which I reduced to proving $4$ points $(B,T,M',J)$ are cyclic where $ABC$ is an acute scalene triangle, $T$ is the intersection of tangents to $(ABC)$ from $B$ and $C$, $M' = AM ∩ (ABC)$ ;($M =$ midpoint $BC$) , and $J = AB ∩ CM'$.
I tried to prove this synthetically but I could not, so I decided to bash. First I tried coordinate but that was too long, so I decided to go with complex because that felt feasible. But then I remembered that I have not yet studied complex numbers (except when I would read EGMO complex numbers for fun, like a storybook). So I opened EGMO and could get complex numbers for everything like how to prove cyclic, intersection, $T$ etc. but Point $M'$ was a problem. I couldn't find how to get it anywhere so I come to MSE.
Please help me, Thanks!
 A: Let $A,B,C$ lie on the unit circle, with complex coordinates $a,b,c$. Since you know EGMO, you should be able to find a proof of the fact that the line through the points $a,b$ on the unit circle has equation
$$z + ab\overline{z} = a+b$$
Now denoting $D = AM \cap (ABC)$ with coordinate $d$, the line $AD$ is given by $z + ad \overline{z} = a+d$.
However you know that the midpoint of $BC$, namely $ \frac{b + c}{2}$ lies on line $AD$, so
$$\frac{b + c}{2} + ad \frac{\overline{b} + \overline{c}}{2} = a+d$$
You can now just solve for $d$ to find the required intersection.
A: 
$AT$ is the symmedian of $\triangle ABC$ through $A$. Hence, $\angle BAT=\angle MAC=\angle M'AC$ and also $\angle CAT=\angle M'AB$.
Now, notice that,  $\angle JAT=\angle BAT=\angle M'AC=\angle M'CT=\angle JCT$ and thus quadrilateral $AJTC$ is cyclic.
Now,  $\angle M'JT=\angle CJT=\angle CAT=\angle BAM'=\angle M'BT$ and therefore quadrilateral $BJTM'$ is cyclic.
P.S. The symmedian through a vertex is the reflection of the median over the internal angle bisector through that vertex. It is provable by euclidean geometry that the symmedian through a vertex goes through the intersection point of the tangents to the circumcircle at the other two vertices.
Check out this link for more:  https://brilliant.org/wiki/symmedian/#:~:text=A%20symmedian%20of%20a%20triangle,D%3D%E2%88%A0CAM.
