Special cases of Apollonius' problem

To construct the perpendicular bisector of the line segment between two points requires two circles, each centered on an endpoint and passing through the other endpoint (operation 2). The intersection points of these two circles (operation 4) are equidistant from the endpoints. The line through them (operation 1) is the perpendicular bisector.

To generate the line that bisects the angle between two given rays requires a circle of arbitrary radius centered on the intersection point P of the two lines (2). The intersection points of this circle with the two given lines (5) are T1 and T2. Two circles of the same radius, centered on T1 and T2, intersect at points P and Q. The line through P and Q (1) is an angle bisector. Rays have one angle bisector; lines have two, perpendicular to one another.

PPP problems generally have a single solution. As shown above, if a circle passes through two given points P₁ and P₂, its center must lie somewhere on the perpendicular bisector line of the two points. Therefore, if the solution circle passes through three given points P₁, P₂ and P₃, its center must lie on the perpendicular bisectors of ${\displaystyle {\overline {\mathbf {P} _{1}\mathbf {P} _{2}}}}$, ${\displaystyle {\overline {\mathbf {P} _{1}\mathbf {P} _{3}}}}$ and ${\displaystyle {\overline {\mathbf {P} _{2}\mathbf {P} _{3}}}}$. At least two of these bisectors must intersect, and their intersection point is the center of the solution circle. The radius of the solution circle is the distance from that center to any one of the three given points.

LLL problems generally offer 4 solutions. As shown above, if a circle is tangent to two given lines, its center must lie on one of the two lines that bisect the angle between the two given lines. Therefore, if a circle is tangent to three given lines L₁, L₂, and L₃, its center C must be located at the intersection of the bisecting lines of the three given lines. 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.

PLL problems generally have 2 solutions. As shown above, if a circle is tangent to two given lines, its center must lie on one of the two lines that bisect the angle between the two given lines. By symmetry, if such a circle passes through a given point P, it must also pass through a point Q that is the "mirror image" of P about the angle bisector. 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 ${\displaystyle {\overline {\mathbf {GT} _{1}}}}$ and ${\displaystyle {\overline {\mathbf {GT} _{2}}}}$ to the solution tangent points T₁ and T₂, are equal to each other and to the product

${\displaystyle {\overline {\mathbf {GP} }}\cdot {\overline {\mathbf {GQ} }}={\overline {\mathbf {GT} _{1}}}\cdot {\overline {\mathbf {GT} _{1}}}={\overline {\mathbf {GT} _{2}}}\cdot {\overline {\mathbf {GT} _{2}}}}$

Thus, the distances ${\displaystyle {\overline {\mathbf {GT} _{1-2}}}}$ are both equal to the geometric mean of ${\displaystyle {\overline {\mathbf {GP} }}}$ and ${\displaystyle {\overline {\mathbf {GQ} }}}$. From G and this distance, the tangent points T₁ and T₂ can be found. Then, the two solution circles are the circles that pass through the three points (P, Q, T₁) and (P, Q, T₂), respectively.

PPL problems generally have 2 solutions. 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 ${\displaystyle {\overline {PQ}}}$ with l. In that case, the sole solution circle is the circle that passes through the three points P, Q and T.

If the line m is not parallel to the given line l, then it intersects l at a point G. By the power of a point theorem, the distance from G to a tangent point T must equal the geometric mean

${\displaystyle {\overline {\mathbf {GT} }}\cdot {\overline {\mathbf {GT} }}={\overline {\mathbf {GP} }}\cdot {\overline {\mathbf {GQ} }}}$

Two points on the given line L are located at a distance ${\displaystyle {\overline {\mathbf {GT} }}}$ from the point G, which may be denoted as T₁ and T₂. The two solution circles are the circles that pass through the three points (P, Q, T₁) and (P, Q, T₂), respectively.

The two circles in the Two points, one line problem where the line through P and Q is not parallel to the given line l, can be constructed with compass and straightedge by:

• Draw the line m through the given points P and Q .
• The point G is where the lines l and m intersect
• Draw circle C that has PQ as diameter.
• Draw one of the tangents from G to circle C.
• point A is where the tangent and the circle touch.
• Draw circle D with center G through A.
• Circle D cuts line l at the points T1 and T2.
• One of the required circles is the circle through P, Q and T1.
• The other circle is the circle through P, Q and T2.

CPP problems generally have 2 solutions. Consider a circle centered on one given point P that passes through the second point, Q. Since the solution circle must pass through P, inversion in this circle transforms the solution circle into a line lambda. The same inversion transforms Q into itself, and (in general) the given circle C into another circle c. Thus, the problem becomes that of finding a solution line that passes through Q and is tangent to c, which was solved above; there are two such lines. Re-inversion produces the two corresponding solution circles of the original problem.

CLP problems generally have 4 solutions. The solution of this special case is similar to that of the CPP Apollonius solution. Draw a circle centered on the given point P; since the solution circle must pass through P, inversion in this circle transforms the solution circle into a line lambda. In general, the same inversion transforms the given line L and given circle C into two new circles, c₁ and c₂. 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.

CCP problems generally have 4 solutions. The solution of this special case is similar to that of CPP. Draw a circle centered on the given point P; since the solution circle must pass through P, inversion in this circle transforms the solution circle into a line lambda. In general, the same inversion transforms the given circle C₁ and C₂ into two new circles, c₁ and c₂. 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.

CLL problems generally have 8 solutions. This special case is solved most easily using scaling. The given circle is shrunk to a point, and the radius of the solution circle is either decreased by the same amount (if an internally tangent solution) or increased (if an externally tangent circle). 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. This shrinking of the given circle to a point reduces the problem to the PLL problem, solved above. In general, there are two such solutions per quadrant, giving eight solutions in all.

CCL problems generally have 8 solutions. The solution of this special case is similar to CLL. The smaller circle is shrunk to a point, while adjusting the radii of the larger given circle and any solution circle, and displacing the line parallel to itself, according to whether they are internally or externally tangent to the smaller circle. This reduces the problem to CLP. Each CLP problem has four solutions, as described above, and there are two such problems, depending on whether the solution circle is internally or externally tangent to the smaller circle.