Produce AB to at E, construct a point C on such that .
Construct the circle ABC.
By Lemma 1, the circle ABC touches at C. ■
Problem 3 [LLL] Given 3 straight lines, construct a circle touching the 3 given straight lines. (It is necessary that no tow of the 3 straight lines are parallel to each other.)
(Remark: the proof is the same as Elements IV.4)
Let B, D be 2 intersections.
Now bisect angle B and angle D. The 2 angle bisectors intersect at E.
Fall from E (3 perpendiculars) to the 3 given straight lines at A, F, C.
Construct the circle AFC.
Since , so EA = EF.
Also, , so EF = EC.
Therefore, EA = EF = EC and E is the center of the circle AFC, which touches the 3 given straight lines at A, F, C. ■
Lemma 2 To draw a line passing through a given point such that it intersects 2 given lines at equal angles.
(case 1) (case 2)
Given a point A and 2 straight lines BC, DE.
If BC is parallel to DE, then fall perpendicular to BC and DE, intersecting both at right angles.
If BC is not parallel to DE, then the 2 lines intersect (at F).
Problem 4 [PLL] Given 2 straight lines and a point, construct a circle passing through the given point and touching the 2 given straight lines.
Given a point A and 2 lines BC and DE.
By Lemma 2, through A draw a line intersecting BC, DE at H, I respectively with equal angles.
So, we have the point K, the intersection of angle bisector and the perpendicular from A, and the point L, the reflection of A about the angle bisector (AK = KL).
Construct a point M on BC such that .
Construct the circle ALM.
Obtain the centre N, fall perpendicular to DE at O.
(Prove the circle ALM touches BC and DE at M and O respectively, i.e. NM = NO)
Since (SAS: NK = NK, , NH = NI),
Therefore and NH = NI.
(ASA), it follows that MN = NO. ■
Problem 5 [CLL] Given a circle and 2 straight lines, construct a circle touching the given circle and the 2 given straight lines.
Given a circle with centre A and two straight line ZC, DB.
Translate ZC, DB parallel to XH, FG respectively by the radius of the given circle. (Viete’s word: perp. to AZ, AD resp).
The dotted circle E shall passes through A and touches both lines XH and FG.
By Problem 4 [PLL], construct the circle passing A and touching XH, FG.
(The proof) Simply reverse the above steps. ■
Lemma 3 If two circles are mutually secant (intersect at two distinct points), one’s diameter must not pass through the center of the other circle.
Given that the circles CEB and CED are mutually secant (not mutually tangent), they must intersect at two points, C and E, say.
Let A be the centre of circle CEB.
Produce CA to meet the circle A at B and the other circle at D.
Claim that the line CABD does not pass through the circle of the circle CED.
Join EC, EB, ED.
So, in the semi-circle, .
Plus or minus the angle , therefore .
Hence CABD is not the diameter of the circle CED, and therefore does not pass through the centre. ■
Lemma 4 If two circles intersect (not mutually tangent), from an intersecting point, draw a line cutting the two circles, the two arcs of the cutting segments are not similar (with respect to their circles).
Given two circles CEB and CED are mutually secant, intersect at C, E.
The line CBD cuts the circle CEB at C, B and cuts the circle CED at C, D.
Claim: arcs CB and CD are not similar.
If CB passes through the center of CEB (A),
Then CB does not pass through the center of the circle CED (by Lemma 3).
Now CB is a diameter of circle CED, but not a diameter of circle CED; therefore CE, CD are not similar arcs.
If CB does not pass through the center of CEB (A),
produce CA to meet the circle CEB ar G and circle CED at F. Join BG, DF.
B, F are points on the same side; C, G are points nearer to M.
Join BK, CK, FL, LG. And so triangles BKC, FLC are formed.
Therefore by construction, , , .
Similarly, , , .
So, the third angles must be equal, i.e. .
Hence segments BC, FG are similar. ■
(M can be inside or outside KL.)
Lemma 7 Given two circles ABCD, EFGH, the line joining the centers KL cuts the circles at A,D, E, H, where AD is a diameter of the first circle and EH is a diameter of the second, M is constructed by Lemma 6. MGFCB cuts the first circle at B, C; cuts the second circle at F, G (hence similar arcs). In the first circle, A, B are farther points and C, D are nearer to M; in the second circle, F, E are farther and G, H are nearer.
Fall perpendiculars from K to BC (to R); from L to FG (to S). Join KC, LG.
BC and FG are similar.
By the half, RC and SG are similar.
Therefore, and .
Even, and finally the subtended chords are parallel .
But . (Remark: by AAA)
In circle EFGH, ,
Therefore , hence . ■
Problem 9 [CCP] Given two circles and a point, construct a circle passing through the given point and touching the two given circles.
(Remark: H does not necessarily lie on the desired circle.)
Given two circles ABCD (center K), EFGH (center L) and a point I.
Join KL and cuts circles such that AD, EH are diameters of the first and second circles respectively.
Also obtain the homothetic center M.
Join MI and construct the point N on MI such that .
Construct the circle passing through I, N and touching the circle ABCD. (by Problem 8 PPC)
Let the point of contact be B and join BM, cutting the circles ABCD and EFGH at B, C and F, G respectively.
Since (by Lemma 7)
By construction, (N, I, B, G concyclic)
But G also lies on the circle EFGH, so the two circles EFGH and IBGN are either mutually secant or mutually tangent to each other.
Since the two circles BNI and BCD are mutually tangent to each other. (Remark: Here B becomes the homothetic center.)
For BG and BC are similar, and FG and BC are similar; thus FG and BG are similar.
Therefore, the two circles EFGH and IBGN are mutually tangent to each other.
In conclusion, the circle IBG is the desired one which passes through the given point I and touching the two given circles ABD, EGH at B, G respectively. ■
Problem 10 [CCC] Given three circles, construct a circle touching the three given circles.
Given a first circle A , second B, third D (remark: with radii respectively)
Construct a circle centered at D with radius .
Construct a circle centered at B with radius .
Lastly, through A construct a circle AGF which touches the two newly constructed circles at G, F; and that this circle has a center E.
Increase the radius by , the circle HLM then satisfies the requirement.