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`Elementary Geometry Miscellaneous Propositions  The problems known as the Tangencies   Case I    Cor   Case II   Case III   Case IV   Case V    Cor Sources and References`

# Elementary Geometry

## Miscellaneous Propositions

### The problems known as the Tangencies

937 Given in position any three of the following nine data: viz., three points, three straight lines, and three circles, it is required to describe a circle passing through the given points and touching the given lines or circles. The following five principal cases occur. #### Case I

938 Given two points, 𝐴, 𝐵, and the straight line 𝐶𝐷.
Analysis: let 𝐴𝐵𝑋 be the required circle, touching 𝐶𝐷 in 𝑋. Therefore 𝐶𝑋2=𝐶𝐴⋅𝐶𝐵III. 36 Hence the point 𝑋 can be found, and the centre of the circle defined by the intersection of the perpendicular to 𝐶𝐷 through 𝑋 and the perpendicular bisector of 𝐴𝐵. There are two solutions.
Otherwise, by (926), making the ratio one of equality, and 𝐷𝑂 the given line.
##### Cor
The point 𝑋 thus determined is the point in 𝐶𝐷 at which the distance 𝐴𝐵 subtends the greatest angle. In the solution of (941) 𝑄 is a similar point in the circumference 𝐶𝐷. (III. 21 & I.16

#### Case II

939 Given one point 𝐴 and two straight lines 𝐷𝐶, 𝐷𝐸.
In the last figure draw 𝐴𝑂𝐶 perpendicular to 𝐷𝑂, the bisector of the angle 𝐷, and make 𝑂𝐵=𝑂𝐴, and this case is solved by Case I.

#### Case III

940 Given the point 𝑃, the straight line 𝐷𝐸, and the circle 𝐴𝐶𝐹.
Analysis: Let 𝑃𝐸𝐹 be the required circle touching the given line in 𝐸 and the circle in 𝐹.
Through 𝐻, the centre of the given circle, draw 𝐴𝐻𝐶𝐷 perpendicular to 𝐷𝐸. Let 𝐾 be the centre of the other circle. Join 𝐻𝐾, passing through 𝐹, the point of contact. Join 𝐴𝐹, 𝐸𝐹, and 𝐴𝑃, cutting the required circle in 𝑋. Then ∠𝐷𝐻𝐹=𝐿𝐾𝐹I.27 therefore 𝐻𝐹𝐴=𝐾𝐹𝐸 (the halves of equal angles); therfore 𝐴𝐹, 𝐹𝐸 are in the same straight line. Then, because 𝐴𝑋⋅𝐴𝑃=𝐴𝐹⋅𝐴𝐸, (III.36) and 𝐴𝐹⋅𝐴𝐸=𝐴𝐶⋅𝐴𝐷 by similar triangles, therefore 𝐴𝑋 can be found.
A circle must then be described through 𝑃 and 𝑋 to touch the given line, by Case I. There are two solutions with exterior contact, as appears from Case I. These are indicated in the diagram. There are two more in which the circle 𝐴𝐶 lies within the described circle. The construction is quite analogous, 𝐶 taking the place of 𝐴.

#### Case IV

941 Given two points 𝐴, 𝐵 and the circle 𝐶𝐷.
Draw any circle through 𝐴, 𝐵 cutting the required circle in 𝐶, 𝐷. Draw 𝐴𝐵 and 𝐷𝐶, and let them meet in 𝑃. Draw 𝑃𝑄 to touch the given circle. Then, because 𝑃𝐶⋅𝑃𝐷=𝑃𝐴⋅𝑃𝐵=𝑃𝑄2III.36 and the required circle is to pass through 𝐴, 𝐵; therefore a circle drawn through 𝐴, 𝐵, 𝑄 must touch 𝑃𝑄, and therefore the circle 𝐶𝐷, in 𝑄 (III.37), and it can be described by Case I. There are two solutions corresponding to the two tangents from 𝑃 to the circle 𝐶𝐷.

#### Case V

942 Given one point 𝑃, and two circles, centres 𝐴 and 𝐵. Analysis: Let 𝑃𝐹𝐺 be the required circle touching the given ones in 𝐹 and 𝐺. Join the centres 𝑄𝐴, 𝑄𝐵. Join 𝐹𝐺, and produce it to cut the circles in 𝐸 and 𝐻, and the line of centres in 𝑂. Then, by the isosceles triangles, the four angles at 𝐸, 𝐹, 𝐺, 𝐻 are all equal; therefore 𝐴𝐸, 𝐵𝐺 are parallel, and so are 𝐴𝐹, 𝐵𝐻; therefore 𝐴𝑂∶𝐵𝑂∷𝐴𝐹∶𝐵𝐻, and 𝑂 is a centre of similitude for the two circles. Again, ∠𝐻𝐵𝐾=2𝐻𝐿𝐾, and 𝐹𝐴𝑀=2𝐹𝑁𝑀 (III. 20); therefore 𝐹𝑁𝑀=𝐻𝐿𝐾=𝐻𝐺𝐾 (III. 21); therefore the triangles 𝑂𝐹𝑁, 𝑂𝐾𝐺 are similar; therefore 𝑂𝐹⋅𝑂𝐺=𝑂𝐾⋅𝑂𝑁; therefore, if 𝑂𝑃 cut the required circle in 𝑋, 𝑂𝑋⋅𝑂𝑃=𝑂𝐾⋅𝑂𝑁. Thus the point 𝑋 can be found, and the problem is reduced to Case IV.
Two circles can be drawn through 𝑃 and 𝑋 to touch the given circles. One is the circle 𝑃𝐹𝑋. The centre of the other is at the point where 𝐸𝐴 and 𝐻𝐵 meet if produced, and this circle touches the given ones in 𝐸 and 𝐻. 943 An analogous construction, employing the internal centre of similitude 𝑂′, determines the circle which passes through 𝑃, and touches one given circle externally and he other internally. See (1047-9).
The centres of similitude are the two points which divide the distanc between the centres in the ratio of the radii. See (1037). 944
##### Cor
The tangents from 𝑂 to all circles which touch the givn circles, either both externally or both internally, are equal.
For the square of the tangent is always equal to 𝑂𝐾⋅𝑂𝑁 or 𝑂𝐿⋅𝑂𝑀. 945 The solutions for the cases of three given straight lines or three given points are to be found in Euc. IV. Props, 4,5. 946 In the remaining cases of the tangencies, straight lines and circles alone are given. By drawing a circle concentic with the required one through the centre of the least given circle, the problem can always be made to depend upon one of the rpeceding cases; the centre of the least circle becoming one of the iven points.

## Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive

ID: 210900019 Last Updated: 9/19/2021 Revision: 0 Ref: References

1. Hilbert, D. (translated by Townsend E.J.), 1902, The Foundations of Geometry
2. Moore, E.H., 1902, On the projective axioms of geometry
3. Fitzpatrick R. (translated), Heiberg J.L. (Greek Text), Euclid (Author), 2008, Euclid's Elements of Geometry  Home 5

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