- Special cases of Apollonius' problem
In

Euclidean geometry ,Apollonius' problem is to construct all the circles that are tangent to three given circles.**Limiting cases of Apollonius' problem**are those in which at least one of the given circles is a point or line, i.e., is a circle of zero or infinite radius. There are nine types of such limiting cases of Apollonius' problem, namely, to construct the circles tangent to:# three points (denoted PPP, generally 1 solution)

# three lines (denoted LLL, generally 4 solutions)

# one line and two points (denoted LPP, generally 2 solutions)

# two lines and a point (denoted LLP, generally 2 solutions)

# one circle and two points (denoted CPP, generally 2 solutions)

# one circle, one line, and a point (denoted CLP, generally 4 solutions)

# two circles and a point (denoted CCP, generally 4 solutions)

# one circle and two lines (denoted CLL, generally 8 solutions)

# two circles and a line (denoted CCL, generally 8 solutions)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 given parallel lines and one given circle.

**Historical introduction**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 be to show that at least two of the angles of an isoceles 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 iff ("something about no higher than square roots are taken"). Therefore, it is important to determine whether an object can be constructed with compass and straightedge and, if so, how it may be constructed.Euclid developed numerous constructions using compass and straightedge. 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 Irish knot-work, can be designed using only Euclidean constructions. However, some geometrical constructions are simply impossible with those tools, including the heptagon and trisecting an angle.

Apollonius contributed a set of constructions, namely, finding the circles that are tangent to three geometrical elements simultaneously, where the "elements" may be a point, line or circle.

**Rules of Euclidean constructions**In Euclidean constructions, only five basic operations are allowed:

# Draw a line through two given points

# Draw a circle through a given point with a given center

# Find the intersection point of two given lines

# Find the intersection points of two given circles

# Find the intersection points of a given line and a given circleThe initial elements in a geometric construction are called the "givens", such as a given point, a given line or a given circle.

**Example 1: Perpendicular bisector**A common task is to construct the perpendicular bisector of the line segment between two endpoints. To do this, one draws two circles, each centered on an endpoint and passing through the other endpoint (operation #2). The intersection points of these two circles may be found (operation #4), and both are equidistant from the endpoints. The line drawn through them (operation #1) is the perpendicular bisector, being the locus of points equidistant from both endpoints.

**Example 2: Angle bisector**Another simple problem is to generate the line that bisects the angle between two given rays. We begin by drawing a circle of arbitrary radius centered on the intersection point P of the two given lines (operation #2). The intersection points T1 and T2 of this circle with the two given rays may be found (operation #5). Two circles of the same radius, centered on T1 and T2, intersect at two points, P and Q. The line drawn through P and Q (operation #1) is an angle bisector. For rays, there is only one angle bisector; for lines, however, there are two, perpendicular to one another.

**Preliminary results**A few basic results are helpful in solving different special cases of Apollonius' problem. The most basic is that a line and a point can be thought of as circles of infinitely large and infinitely small radius, respectively. A circle is tangent to a point if it passes through the point, and tangent to a line if they intersect at a single point

**P**and if the line is perpendicular to a radius drawn from the circle's center to**P**.**Circles tangent to two given points**Must lie on the perpendicular bisector

**Circles tangent to two given lines**Must lie on the angle bisector

**Tangent line to a circle from a given point**Draw semicircle centered on the midpoint between the center of the circle and the given point

**Power of a point and the harmonic mean****Radical axis of two circles**set of points of equal tangents, or more generally, equal power.

**Circle inversion**circles into lines, circles into circles

**caling two circles while maintaining their tangency**If two circles are "internally" tangent, they remain so if their radii are increased or decreased by the same amount. Conversely, if two circles are "externally" tangent, they remain so if their radii are changed by the same amount in opposite directions, one increasing and the other decreasing.

**Types of solutions****Type 1: Three points (PPP, generally 1 solution)**As shown above, if a circle passes through two given points

**P**_{1}and**P**_{2}, its center must lie somewhere on the perpendicular bisector line of the two points. Therefore, if the solution circle passes through three given points**P**_{1},**P**_{2}and**P**_{3}, its center must lie on the perpendicular bisectors of $overline\{mathbf\{P\}\_\{1\}mathbf\{P\}\_\{2$, $overline\{mathbf\{P\}\_\{1\}mathbf\{P\}\_\{2$ and $overline\{mathbf\{P\}\_\{1\}mathbf\{P\}\_\{2$. 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.**Type 2: Three lines (LLL, generally 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 "L"

_{1}, "L"_{2}, and "L"_{3}, 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 as follows. Choose one of the three intersection points**P**between the given lines; draw 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.**Type 3: One point, two lines (PLL, generally 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 a 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 will pass through both**P**and**Q**, and theirradical axis is the line connecting those two points. Consider the 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 $overline\{mathbf\{GT\}\_\{1$ and $overline\{mathbf\{GT\}\_\{1$ to the solution tangent points**TT"'_{1}and_{2}, are equal to each other and to the product:$overline\{mathbf\{GP\; cdot\; overline\{mathbf\{GQ\; =\; overline\{mathbf\{GT\}\_\{1\; cdot\; overline\{mathbf\{GT\}\_\{1\; =overline\{mathbf\{GT\}\_\{2\; cdot\; overline\{mathbf\{GT\}\_\{2$

Thus, the distances $overline\{mathbf\{GT\}\_\{1-2$ are both equal to the

harmonic mean of $overline\{mathbf\{GP$ and $overline\{mathbf\{GQ$. From**G**and this distance, the tangent points**T**_{1}and**T**_{2}can be found. Then, the two solution circles are the circles that pass through the three points (**P**,**Q**,**T**_{1}) and (**P**,**Q**,**T**_{2}), respectively.**Type 4: Two points, one line (PPL, generally 2 solutions)**A line "l" may be drawn through the given points

**P**and**Q**. If the line 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 $overline\{PQ\}$ 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 "l" 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 theharmonic mean :$overline\{mathbf\{GT\; cdot\; overline\{mathbf\{GT\; =\; overline\{mathbf\{GP\; cdot\; overline\{mathbf\{GQ$

Two points on the given line "L" are located at a distance $overline\{mathbf\{GT$ from the point

**G**, which may be denoted as**T**_{1}and**T**_{2}. The two solution circles are the circles that pass through the three points (**P**,**Q**,**T**_{1}) and (**P**,**Q**,**T**_{2}), respectively.**Type 5: One circle, two points (CPP, generally 2 solutions)**Draw 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.**Type 6: One circle, one line, one point (CLP, generally 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, "c"_{1}and "c"_{2}. 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.**Type 7: Two circles, one point (CCP, generally 4 solutions)**The solution of this special case is likewise 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 circle "C"_{1}and "C"_{2}into two new circles, "c"_{1}and "c"_{2}. 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.**Type 8: One circle, two lines (CLL, generally 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, which was solved above. In general, there are two such solutions per quadrant, giving eight solutions in all.

**Type 9: Two circles, one line (CCL, generally 8 solutions)**The solution of this special case is similar to that of the CCL Apollonius problem, immediately above. The smaller circle is again shrunk to a point, while adjusting the radii of the larger given circle and any solution circle, according to whether they are internally or externally tangent to the smaller circle. This reduces the problem to the PLC Apollonius problem, which was solved above. Each PLC 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.

**pecial cases with no solutions**An Apollonius problem is impossible if the given circles are "nested", i.e., if one circle is completely enclosed within a particular circle and the remaining circle is completely excluded. This follows because any solution circle would have to cross over the middle circle to move from its tangency to the inner circle to its tangency with the outer circle. This general result has several special cases when the given circles are shrunk to points (zero radius) or expanded to straight lines (infinite radius). For example, the CCL problem has zero solutions if the two circles are on opposite sides of the line since, in that case, any solution circle would have to cross the given line non-tangentially to go from the tangent point of one circle to that of the other.

**ee also***

Problem of Apollonius

*Compass and straightedge constructions **References***

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**External links**

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