Line And Two Points In The Same Half-Plane

Steps

0. • Two points A and B and a line p are given.
1. • Choose the circle ω to perform the circular inversion so that its center is one of the given points. The circle may pass through the second given point. This will make the second point self-conjugate.
2. • The image of the line p is the circle p'.
3. • Since point B is the center of the circle ω, it will appear at infinity. Therefore, the image of the circles we are looking for are the tangents of the circle p' drawn from point A.
4. • By representing the tangents in the circular inverse, we obtain the solution circles.
5. • The problem has two solutions.

Steps

0. • Two points A and B and a line p are given.
1. • The set of points that have the same distance from a line and a point is a parabola. The focus of the parabola is the given point and the directrix is the given line. Draw two parabolas, one for each given point. We could also use the axis of the line segment AB, on which points with the same distance from points A and B lie.
2. • The intersections of the two parabolas are the centres of the circles we are looking for.
3. • The solution is the two circles.

Steps

0. we have the line p and the points A, B (in brackets are names of the objects in the situation, that B is further away from the line p than A)
1. draw the line g passing through the points A and B, name the intersection gp E, construct the perpendicular line h to the line g passing through the point B and the perpendicular line i to the line g passing through the point A and the line
2. on the midpoint between A and E, we place the point F, which is the centre of the circle c with radius the size of the line AF
3. at the intersections of circles c (f) and h (i) are points C D (J)
4. draw a circle d (q) centered at E with radius of line size ED (EJ)
5. at the intersections of the line p and the circle d (q) there are points G (K) and H (L), which are tangent points to the resulting circles
6. according to the three points we are able to draw the resulting circles k1 given by the points BAL a k2 given by the points BAK

Steps

1. Draw the line AB and its axis, because the solution will lie at the same distance from both points. Where the axis intersects the line p from the problem, the centre of equilibrium will lie, call it E.
2. Draw a circle c centered at any point on the axis of line AB, which is also tangent to line p from the problem. The tangent point on the line p lies on its perpendicular, which passes through the centre of the circle. This circle is a representation of the resulting circle in the equilateral plane.
3. Draw the lines passing through the centre of the equidistance, i.e. the point E, and the points A and B from the problem. The intersections of these lines with the circle c are representations of points A and B. Let us call F, G, H and I.
4. Draw lines through the center of the circle c and through the image of points A (that is, F, G) and B (H and I).
5. From each of these lines, we draw a parallel line that passes through the corresponding point in the image, since we know that the identity preserves parallelism. The intersections of these parallel lines will be the centers of the resulting circles.
6. The intersections of these parallels are the centers of the two resultant circles.
7. The resultant circles of the line, point, point problem solved using the equiangularity.