Find the equation of the line through the point $(1,-1)$ Find the equation of the line through the point $(1,-1)$ which cuts off a chord of length $4\sqrt {3}$ from the circle $x^2+y^2-6x+4y-3=0$
My Attempt:
Let the equation of line be $y=mx+c$ 
It passes through $(1,-1)$ so
$$-1=m+c$$ 
$$c=-1-m$$
 A: $$(x-3)^2+(y+2)^2=3+9+4=4^2$$
Let the center be $O(3,-2)$
Let one of the intersections of the line with the circle be $P$
So, $|OP|=4$
Let the perpendicular from the center to the line intersect at $Q$
Using The perpendicular from the center of a circle to a chord bisects the chord (Proof)
$|PQ|=2\sqrt3$
$$OP^2=PQ^2+OQ^2\iff OQ^2=4^2-(2\sqrt3)^2\implies|OQ|=?$$
Now the equation of the line $$mx-y-1-m=0$$
The perpendicular distance from $O$ to the line $$\dfrac{|3m+2-1-m|}{\sqrt{1+m^2}}=|OQ|$$
Square both sides to find $m$
A: Hint or Plan: From where you left off, the equation of the line is: $y = mx-m-1$, and find the $2$ points that it intersects the circle via the equation: $(x-3)^2+(mx-m-1+2)^2=4^2$. You then solve for $x$ and using quadratic formula to get $2$ solutions of $x$ in term of $m$, then reuse the equation of the line to get the coresponding $y$ values for each solution $x$. Then use the distance formula to find the distance between the two points and set them equal to $4\sqrt{3}$ and solve for $m$.
A: The points of $\mathbb{R}^2$ that are both in the circle you're given and in a generic line (passing through that point) are given by the system of equations:
$$
\begin{cases}
x^2+y^2-6x+4y-3 = 0 \\
(y+1) = m(x-1)
\end{cases}
$$
Now we wish our two solutions to be described by this equation:
$${(x_1 - x_2)}^2 + {(y_1 - y_2)}^2 = (4\sqrt{3})^2$$
Begin by substituting.
$$x^2+(m(x-1) - 1)^2 -6x+4(m(x-1) - 1)-3 = 0 \\
x^2 + m^2(x-1)^2 - 2m(x-1) + 1 - 6x + 4(m(x-1) - 1)-3 = 0 \\
x^2 + m^2x^2 - 2m^2x + m^2 + 2mx + 2m - 6x + 6 = 0
$$
Note that this equation has your points in it, given the right $m$. This equation is just a (slightly complicated) parabola. This means you can find the roots of this equation, for any $m$.
Once you've found the roots for the equation, you can plug them in the third equation I gave you.
I'm sure there is a much easier method to solve this. I'll come back to you as soon as possible.
EDIT: Here's what I thought.
Consider the circumference in its normal form:
$$(x - 3)^2 + (y + 2)^2 = 16 $$
Now translate it in the origin and scale it down evenly so that it has radius 1. Where is a possible chord of length $\sqrt{3}$ located?
It is twice the half chord spanned by the angle $\pi/3$. The distance of this chord from the center is half the radius; this comes from trigonometry. Notice that this line, in the reference frame of the original circle, is exactly the line you're looking for, and it must pass through the point.
We now have everything we need.
Draw a circle with half the radius, concentric with the original. 
$$ (x-3)^2 + (y+2)^2 = 4$$
Now notice that your point is outside of this circle. Since we defined our chord to be both passing through the point and being distant half the radius from the center, your (two chords, because there are two that give you the same results) chord is exactly the line (lines) tangent to the smaller circle and passing through $(1, -1)$.
Formulas for these lines can be easily found on Google. You then just need to substitute the final equation of the line in the first system I wrote, and solve for $x$ and $y$.
If these guidelines are not enough, I'll provide the full solution in another edit.
