Hello, StaryNight!
Given you have the equation of a circle and a point inside it, how do you
calculate the equation of a chord that has that point as its midpoint?
To give an example: let circle $\displaystyle C$ be: .$\displaystyle x^2 + y^2 2x + 4y 11 \:= \:0$
Let the desired midpoint be $\displaystyle A(2,1)$ Code:


 * * *
*  * *
*+**
*  * * *
 o A(2,1)
*  C * * *
*  o * *
*  (1,2) *

*  *
*  *
*  *
 * * *

We find that the equation of the circle is: .$\displaystyle (x1)^2 + (y+2)^2 \:=\:16$
. . It has center $\displaystyle C(1,2)$ and radius 4.
The radius that bisects a chord is perpendicular to the chord.
The slope of $\displaystyle CA$ is: .$\displaystyle m_1 \:=\:\frac{\text{}1(\text{}2)}{21} \;=\;1$
. . Hence, the slope of the chord is: .$\displaystyle m_2 \:=\:1$
Now write the equation of the line through $\displaystyle A(2,1)$ with slope $\displaystyle m_2 = 1.$