I'll provide a quickly-drawn representation of what the problem is. Basically, there is a line $$l: y=-x+b$$ and there are 2 known points on it: $$A = (-6,8)$$ and $$B = (-2,4)$$ The line in question (let's name it k, it's the red one) passes through point A. Additionally, the distance between l and point C, located on k, as shown on the image, is 2. Is it possible to get the equation of k with just this amount of information?
On
Imagine that segment of length $2$ on a pivot at point $B$. As you rotate this piece around $B$, you change the position of the point $C$ on the other end. In particular, $C$ takes a circular path, and is represented by a circular function (trigonometric function) of the angle between the segment and the line $l$, which we'll call $\theta$. In order for the line $k$ to be independent of the angle, the path of $C$ must be a set of points which make a line with $A$; otherwise, there will not be a way to draw $k$ as a straight line through all the points. Since this is not the case, the line $k$ is a non-constant function of the angle of the pivot, i.e. $k = f(\theta)$.
If the condition is that the shortest distance between $B$ and any point on $k$ is $2$, we can "uniquely" define $k$ if and only if the distance from $A$ to $B$ is $2$ or more. In fact, there are two symmetric solutions where $k$ is below and above $l$ to the right of $A$. For the given $A$ and $B$, this distance is $4\sqrt{2} \geq 2$, so we can find the line $k$; it will be the tangent line of the circle centered at $B$ with radius $2$ which passes through $A$; we'll call the tangent point $T$. The angle of $ATB$ is a right angle because $BT$ is a radius and $TA$ is a tangent. The length $BT$ is $2$ and the length $AB$ is $4\sqrt{2}$, so the length $AT$ must be exactly $2\sqrt{7}$. Then, consider the circle of radius $2\sqrt{7}$ centered at $A$. At the intersections of the two circles, the tangent line of one circle is the radius of the other, so the tangent lines will intersect at a right angle. This means that those intersections are the two solutions for the point $T$. To find the intersection of two circles, we solve the simultaneous equations $$ (x+2)^2+(y-4)^2 = \ 4 \\ (x+6)^2+(y-8)^2 = 28 $$ The solutions are the two points $$T_{+,-} = \left(\frac{-5\pm\sqrt{7}}{2},\frac{9\pm\sqrt{7}}{2}\right)$$
Thus, the equation for $k$ is $$y = \left(\frac{-7\pm\sqrt{7}}{ \ \ \ 7\pm\sqrt{7}}\right) x + \left(\frac{14\pm14\sqrt{7}}{7\pm\sqrt{7}}\right)$$
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To find the contact points of the tangents from $A(-6,8)$ to the circle with center $B(-2,4)$ and radius $2$, given by $(x+2)^2+(y-4)^2=4$, first write down the equation of the polar of $A$ in respect of this circle: $$(x+2)(-6+2)+(y-4)(8-4)=4\iff y=x+7.$$ Plug this in the circle’s equation to get the contact points $(-2.5\pm\sqrt7/2,4.5\pm\sqrt7/2)$.
No. For the line to be uniquely specified, you would need to uniquely specify the triangle ABC. You only have two side lengths, which is not enough to determine it.
The length AC, angle BAC, or angle ABC would be sufficient. Angle ACB could work, but there might be two solutions.
If ACB is intended to be a right angle, then there is a unique solution for line $k$. This solution can be found from the fact that you know the side lengths of triangle ABC from the Pythagorean theorem, and thus can set up a pair of quadratic equations for the coordinates of C using the distance formula.