Hello!

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

`(x-3)^2+(y+2)^2=4^2.`

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:

`(x-3)^2+(mx-(m-1))^2=16,`

`x^2(1+m^2)-2x(3+m(m-1))+(m-1)^2-7=0.`

This is a quadratic equation for x and it has two solutions. The square of the difference of them is

`(dx)^2=4/(1+m^2)^2 (12m^2-4m+15).`

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.`

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