The equation of a line is `y=m(x-1)-1,` except for the vertical line x=1. The given equation is actually an equation of a circle, it is equivalent to
The radius of this circle is 4 and the diameter is 8, which is greater than `4sqrt(3),` so there must be two solutions.
Substitute y from the equation of a line into the equation of the circle:
This is a quadratic equation for x and it has two solutions. The square of the difference of them is
Therefore `(dy)^2=m^2 (dx)^2` and `(dx)^2+(dy)^2=4/(1+m^2) (12m^2-4m+15)` and it must be equal to `(4sqrt(3))^2.` Solving this equation we obtain only one m=3/4. Where is the second? It is the vertical line (check this, it is simple).
The answers: `x=1` and `y=3/4 (x-1) -1.`