Alright, let's tackle your problem step by step, breaking down each scenario with existence checks, equation-solving algorithms, and proofs of impossibility where applicable.

1. Circle Tangent to Three Line Segments ([A₁,B₁], [A₂,B₂], [A₃,B₃]) #

Existence Check #

First, remember that circles tangent to three infinite lines (the lines containing your segments) are the incircle and three excircles of the triangle formed by the lines (if the lines are non-parallel and form a triangle). However, since we're dealing with segments (not infinite lines), we need to verify if these candidate circles actually touch the segments (not just the lines beyond the segments).

For a circle to be tangent to a segment [A,B], one of two conditions must hold:

• The perpendicular from the circle's center to the line AB lands on the segment [A,B], and the distance from the center to the line equals the radius.
• The circle touches one of the endpoints (A or B), and the radius to that endpoint is perpendicular to the segment [A,B].

So first, generate the 4 candidate circles (incircle + 3 excircles) from the three lines. Then for each candidate, check if it satisfies the tangent condition for all three segments. If any candidate passes, a valid circle exists.

If the three lines are parallel:

• A valid circle exists only if the distance between every pair of lines is equal to twice the radius (i.e., all lines are equally spaced, with the middle line exactly halfway between the other two). Otherwise, no such circle can exist (since the radius would need to be two different values simultaneously).

Equation-Solving Algorithm #

Let the circle have center ((x,y)) and radius (r > 0). For each segment [A,B], define:

• Vector (\vec{AB} = (B_x - A_x, B_y - A_y))
• Distance from center to line AB: (d = \frac{|(B_x - A_x)(A_y - y) - (B_y - A_y)(A_x - x)|}{\sqrt{(B_x - A_x)^2 + (B_y - A_y)^2}})
• Projection of (\vec{AO}) (O = center) onto (\vec{AB}): (proj = \frac{(x - A_x)(B_x - A_x) + (y - A_y)(B_y - A_y)}{|\vec{AB}|^2})

For each segment, we have three possible tangent conditions:

1. Tangent to the segment's interior: (d = r) and (0 \leq proj \leq 1) (the perpendicular foot lies on [A,B])
2. Tangent at endpoint A: (\sqrt{(x - A_x)^2 + (y - A_y)^2} = r) and ((x - A_x)(B_x - A_x) + (y - A_y)(B_y - A_y) = 0) (radius is perpendicular to the segment)
3. Tangent at endpoint B: (\sqrt{(x - B_x)^2 + (y - B_y)^2} = r) and ((x - B_x)(B_x - A_x) + (y - B_y)(B_y - A_y) = 0)

This gives (3^3 = 27) possible systems of equations to solve. For each system:

• Solve the three equations for (x, y, r)
• Verify (r > 0) and that all conditions are satisfied (no contradictions)

Impossibility Proof Example #

Suppose we have three parallel segments where the distance between [A₁,B₁] and [A₂,B₂] is 2, between [A₂,B₂] and [A₃,B₃] is 3. For a circle tangent to all three, its radius would need to equal the distance to each line: (r = 2 - r) (from first two lines) implies (r=1), but (r=3 - r) (from last two lines) implies (r=1.5). Contradiction—no such circle exists.

2. Circle Tangent to Two Line Segments and Passing Through a Point P(x₀,y₀) #

Existence Check #

First, find all circles that are tangent to the two infinite lines containing the segments and pass through P. Then filter these candidates to see if they actually tangent the segments (not just the lines).

For non-parallel lines:

• The center of a circle tangent to two lines lies on one of the two angle bisectors of the lines. For each bisector, set up equations where the distance from the center to either line equals the radius, and the distance from the center to P equals the radius. Solve for centers, then check segment tangency.

For parallel lines:

• A valid circle exists only if the distance between the lines is (2r) (so the center is midway between them), and the distance from P to the center equals (r). If the distance between lines is not (2r) (or no such (r) satisfies both conditions), no circle exists.

Additionally, P must be positioned such that it's outside the "wedge" formed by the two lines (for non-parallel lines) or on the correct side relative to the parallel lines to allow a tangent circle.

Equation-Solving Algorithm #

Let the circle have center ((x,y)) and radius (r > 0). The conditions are:

1. Passes through P: ((x - x₀)^2 + (y - y₀)^2 = r^2)
2. Tangent to first segment [A₁,B₁]: One of the three tangent conditions from Scenario 1
3. Tangent to second segment [A₂,B₂]: One of the three tangent conditions from Scenario 1

Steps:

1. Generate all possible combinations of tangent conditions for the two segments (3×3=9 systems)
2. For each system, combine with the "pass through P" equation
3. Solve the system for (x,y,r)
4. Verify (r > 0) and that all tangent conditions are satisfied (e.g., perpendicular foot lies on the segment, or endpoint tangency holds)

Impossibility Proof Example #

Suppose we have two parallel lines 4 units apart, and point P is 1 unit from the top line. For a circle tangent to both lines, (r=2) (since distance between lines is (2r)). But the distance from P to the center (which is midway between the lines, 2 units from each line) is (2 - 1 = 1), which is not equal to (r=2). Contradiction—no such circle exists.

3. Circle Tangent to One Line Segment and Passing Through Two Points P₁(x₁,y₁), P₂(x₂,y₂) #

Existence Check #

The center of any circle passing through P₁ and P₂ lies on the perpendicular bisector of P₁P₂. We need to find a point on this bisector such that:

• The distance from the point to P₁ (or P₂) equals the radius (r)
• The distance from the point to the segment [A,B] equals (r), and the tangent condition holds (foot on segment or endpoint tangency)

If no such point exists on the perpendicular bisector (e.g., the bisector doesn't intersect the set of points at distance (r) from the segment, where (r) equals the distance to P₁), then no circle exists.

Equation-Solving Algorithm #

1. Find the perpendicular bisector of P₁P₂:
   The midpoint of P₁P₂ is (M\left(\frac{x₁+x₂}{2}, \frac{y₁+y₂}{2}\right))
   The slope of P₁P₂ is (m = \frac{y₂ - y₁}{x₂ - x₁}) (if (x₂ ≠ x₁)), so the bisector slope is (-\frac{1}{m}) (or vertical if (x₂ = x₁))
   Bisector equation: ((x₂ - x₁)(x - M_x) + (y₂ - y₁)(y - M_y) = 0)

2. Set up tangent and radius conditions:
   For a center ((x,y)) on the bisector:

   • Radius: (r = \sqrt{(x - x₁)^2 + (y - y₁)^2})
   • Tangent to segment [A,B]: One of the three tangent conditions from Scenario 1 (with (d = r) or endpoint tangency)

3. Solve the system:
   Substitute the bisector equation into the tangent and radius equations, solve for (x,y,r)
   Verify (r > 0) and that the tangent condition is satisfied (e.g., perpendicular foot lies on [A,B])

Impossibility Proof Example #

Suppose P₁ and P₂ are on opposite sides of segment [A,B], and the perpendicular bisector of P₁P₂ intersects [A,B]. Any circle passing through P₁ and P₂ would have its center on the bisector, but the distance from the center to [A,B] would be less than the radius (since the center is on one side of [A,B], and one of P₁/P₂ is on the opposite side, so the radius must be larger than the distance to the segment). Thus, the circle would intersect the segment at two points, not tangent—no such circle exists.

内容的提问来源于stack exchange，提问作者Ivan Z