The nine types of such limiting cases of Apollonius' problem are to construct the circles tangent to: In a different type of limiting case, the three given geometrical elements may have a special arrangement, such as constructing a circle tangent to two parallel lines and one circle.
Like most branches of mathematics, Euclidean geometry is concerned with proofs of general truths from a minimum of postulates.
For example, a simple proof would show that at least two angles of an isosceles triangle are equal.
One important type of proof in Euclidean geometry is to show that a geometrical object can be constructed with a compass and an unmarked straightedge; an object can be constructed if and only if (iff) (something about no higher than square roots are taken).
Examples include: regular polygons such as the pentagon and hexagon, a line parallel to another that passes through a given point, etc.
Many rose windows in Gothic Cathedrals, as well as some Celtic knots, can be designed using only Euclidean constructions.
However, some geometrical constructions are not possible with those tools, including the heptagon and trisecting an angle.
The intersection points of these two circles (operation 4) are equidistant from the endpoints.
To generate the line that bisects the angle between two given rays[clarification needed] requires a circle of arbitrary radius centered on the intersection point P of the two lines (2).
A few basic results are helpful in solving special cases of Apollonius' problem.
Therefore, if the solution circle passes through three given points P1, P2 and P3, its center must lie on the perpendicular bisectors of
The radius of the solution circle is the distance from that center to any one of the three given points.
In general, there are four such points, giving four different solutions for the LLL Apollonius problem.
The radius of each solution is determined by finding a point of tangency T, which may be done by choosing one of the three intersection points P between the given lines; and drawing a circle centered on the midpoint of C and P of diameter equal to the distance between C and P. The intersections of that circle with the intersecting given lines are the two points of tangency.
The two solution circles pass through both P and Q, and their radical axis is the line connecting those two points.
Consider point G at which the radical axis intersects one of the two given lines.
Since, every point on the radical axis has the same power relative to each circle, the distances
If a line m drawn through the given points P and Q is parallel to the given line l, the tangent point T of the circle with l is located at the intersection of the perpendicular bisector of
Thus, the problem becomes that of finding a solution line tangent to the two inverted circles, which was solved above.
There are four such lines, and re-inversion transforms them into the four solution circles of the Apollonius problem.
Thus, the problem becomes that of finding a solution line tangent to the two inverted circles, which was solved above.
There are four such lines, and re-inversion transforms them into the four solution circles of the original Apollonius problem.
Depending on whether the solution circle is increased or decreased in radii, the two given lines are displaced parallel to themselves by the same amount, depending on which quadrant the center of the solution circle falls.