<s>[INST] 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
$$-1=m+c$$ 
$$c=-1-m$$
 [/INST] $$(x-3)^2+(y+2)^2=3+9+4=4^2$$
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 $m$
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