Undergrad Linear Algebra 1 problem, Vector Geometry: Lines

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To find all points on the line L: x = (-3, 1) + t(1,-2) that are 2 units from the point (-3, 1), one approach is to intersect the line with a circle centered at (-3, 1) with a radius of 2. This involves setting up the equation for the circle and substituting the line's parametric equations. Alternatively, one can calculate the distance between the point (-3, 1) and points along the line, adjusting the parameter t until the distance equals 2. The solution yields the coordinates (3 ± 2 / √5, -1 ± 4/√5). This method effectively demonstrates how to apply geometric principles to solve vector geometry problems.
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Given the line L: x = (-3, 1) + t(1,-2) find all x on L that lie 2 units from (-3, 1).
Problem: Given the line L: x = (-3, 1) + t(1,-2) find all x on L that lie 2 units from (-3, 1).

I know the answer is (3 ± 2 / √5, -1 ± 4/√5) but I don't know where to start. I found that if t=2, x= (-5, 5) and the normal vector is (2, 1) but I am not sure if this information is useful or how to use it.
 
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Given a choice of t, can you write down the distance between L(t) and (-3,1)?
 
Student323 said:
Summary:: Given the line L: x = (-3, 1) + t(1,-2) find all x on L that lie 2 units from (-3, 1).

Problem: Given the line L: x = (-3, 1) + t(1,-2) find all x on L that lie 2 units from (-3, 1).

I know the answer is (3 ± 2 / √5, -1 ± 4/√5) but I don't know where to start. I found that if t=2, x= (-5, 5) and the normal vector is (2, 1) but I am not sure if this information is useful or how to use it.
You could intersect the line with a circle around (-3,1) with radius 2. For that set up the equation for the circle and use it for (x,y) on the line.
 
Another way is to step along the line until you get to a point that is ##2## units apart. Calculate the length between ##\binom{-3}{1}## and ##\binom{-3}{1}+t\cdot \binom{1}{-2}## which equals ##2##.
 
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