I have the circumference: $$(x-2)^{2}+(y-1)^{2}=25$$ and two secant lines to it: $$s_{1}\colon3x+4y+c'=0$$ and $$s_{2}\colon3x+4y+c''=0$$ I know that the line $r\colon3x+4y-1=0$ and I also know that $r\parallel s_{1}\parallel s_{2}$. My problem is to find $c'$ and $c''$ such that the secant lines $s_{1}$ and $s_{2}$ make a chord with 8 units length.
In the line $s_{1}$ I did this: $$x=-\frac{4}{3}y-\frac{c'}{3}$$ so I replaced it on the circumference's equation: $$\left(-\frac{4}{3}y-\frac{c'}{3}-2\right)^{2}+(y-1)^{2}=25$$ But I really got a huge equation and I think this is not the best way to find the points in the circumference that belong to the line $s_{1}$. I was trying to do this so I would make the distance between them equal to 8 to find the value of $c'$.
I have to use mathematics from high school, so I'd like to know if there's a better way to do this.
Let $d$ be the perpendicular distance from the circle’s center to the line. The length of the secant is then given by the Pythagorean theorem: $2\sqrt{r^2-d^2}$, where $r$ is the radius of the circle. Plug in the values from this problem and solve for $d$, then use the formula for the distance of a point from a line to determine $c'$ and $c''$.