Hello, milan!
A tangent is drawn to the circle: .$\displaystyle x^2 + y^2  18x + 4y + 35 \:= \:0$ at the point $\displaystyle P(14,3).$
1) Find the equation to the tangent.
I already did . . . it is: .$\displaystyle y + x  17 \:= \:0$
2) Find the equations of the two chords, each of length $\displaystyle 6\sqrt{2}$, which are parallel to the tangent.
We complete the square and get: .$\displaystyle (x9)^2 + (y+2)^2 \:=\:50$
Make sketch . . . Code:
Figure 1
 o
 o
 * * * o
 * * P(14,3)
 * o
 * * *o
  +           *    o
 * * * o
 * C * *
 * (9,2) *

 * *
 * *
 * *
 * * *

The center of the circle is $\displaystyle C(9,\text{}2)$
Its radius is $\displaystyle PC = 5\sqrt{2}$.
The equation of the tangent is: .$\displaystyle y \:=\:x+17$
And its slope is 1.
A new diagram to examine one of the chords.
I rotated the diagram so the chord is horizontal. Code:
Figure 2
* * * _
Q * A 3√2 * P
*    +    *
* *  * _*
*  * 5√2
* *  * *
*     *     *
C
The radius is: $\displaystyle CP = 5\sqrt{2}$
The chord is: $\displaystyle PQ = 6\sqrt{2} \quad\Rightarrow\quad AP = 3\sqrt{2}$
We see that $\displaystyle \Delta PAC$ is a 345 right triangle. .Hence: .$\displaystyle AC = 4\sqrt{2}$
. . That is, the chord is $\displaystyle \tfrac{4}{5}\text{(radius)}$ from the center.
Look at Figure 1.
Point $\displaystyle A$ is $\displaystyle \tfrac{4}{5}$ of the way from $\displaystyle C(9,\text{}2)$ to $\displaystyle P(14,3)$.
. . Hence, point $\displaystyle A$ is at $\displaystyle (13,2)$.
The other chord cross the diameter at $\displaystyle B(5,\text{}6)$.
We know all about the two chords!
. . One contains $\displaystyle A(13,2)$ and has slope $\displaystyle 1$.
. . The other has $\displaystyle B(5,\text{}6)$ and has slope $\displaystyle 1$.
Go for it!