PROP.IX.—Theorem.
PROP.IX.—Theorem.
A point(P)within a circle(ABC), from which more than two equal lines(PA,PB,PC,&c.)can be drawn to the circumference, is the centre.
Dem.—IfPbe not the centre, letObe the centre. JoinOP, and produce it to meet the circle inDandE; thenDEis the diameter, andPis a point in it which is not the centre: therefore [vii.] only two equal lines can be drawn fromPto the circumference; but three equal lines are drawn (hyp.), which is absurd.HencePmustbe the centre.
Or thus:Since the linesAP,BPare equal, the line bisecting the angleAPB[vii.Cor.1] mustpass through the centre: in like manner the line bisecting the angleBPCmust pass through thecentre.Hence the point of intersection of these bisectors, that is, the pointP, must be thecentre.
PROP.X.—Theorem.If two circles have more than two points common, they must coincide.
PROP.X.—Theorem.If two circles have more than two points common, they must coincide.
Dem.—LetXbe one of the circles; and if possible let another circleYhave three points,A,B,C, in common withX, without coinciding with it. FindP, the centre ofX. JoinPA,PB,PC. Then sincePis the centre ofX, the three linesPA,PB,PCare equal to one another.
Again, sinceYis a circle andPa point, from which three equal linesPA,PB,PCcan be drawn to its circumference,Pmust be the centre ofY.HenceXandYare concentric, which[v.]is impossible.
Cor.—Two circles not coinciding cannot have more than two points common. Compare I., Axiomx., that two right lines not coinciding cannot have more than one point common.
PROP.XI.—Theorem.If one circle(CPD)touch another circle(APB)internally at any pointP, theline joining the centres must pass through that point.
PROP.XI.—Theorem.If one circle(CPD)touch another circle(APB)internally at any pointP, theline joining the centres must pass through that point.
Dem.—LetObe the centre ofAPB. JoinOP. I say the centre of the smaller circle is in the lineOP. If not, let it be in any other position such asE. JoinOE,EP, and produceOEthroughEto meet the circles in the pointsC,A. Now sinceEis a point in the diameter of the larger circle between the centre andA,EAis less thanEP[vii.2]; butEPis equal toEC(hyp.), being radii of the smaller circle. HenceEAis less thanEC; which is impossible; consequently the centre of the smaller circle must be in the lineOP. Let it beH; then we seethat the line joining the centres passes through the pointP.
Or thus:SinceEPis a line drawn from a point within the circleAPBto the circumference, but not forming part of the diameter throughE, the circle whose centre isEand radiusEPcuts [vii.,Cor.2]APBinP, but it touches it (hyp.) also inP, which is impossible.Hence the centre of the smaller circleCPDmust be in thelineOP.
PROP.XII.—Theorem.If two circles(PCF,PDE)have external contact at any pointP, the linejoining their centres must pass through that point.
PROP.XII.—Theorem.If two circles(PCF,PDE)have external contact at any pointP, the linejoining their centres must pass through that point.
Dem.—LetAbe the centre of one of the circles. JoinAP, and produce it to meet the second circle again inE. I say the centre of the second circle is in the linePE. If not, let it be elsewhere, as atB. JoinAB, intersecting the circles inCandD, and joinBP. Now sinceAis the centre of the circlePCF,APis equal toAC; and sinceBis the centre of the circlePDE,BPis equal toBD. Hence the sum of the linesAP,BPis equal to the sum of the linesAC,DB; butABis greater than the sum ofACandDB; thereforeABis greater than the sum ofAP,PB—that is, one side of a triangle greater than the sum of the other two–which [I.xx.] is impossible. Hence the centre of the second circle must be in the linePE. Let it beG, and we see thatthe line through the centres passes through the pointP.
Or thus:SinceBPis a line drawn from a point without the circlePCFto its circumference, and when produced does not pass through the centre, the circle whose centre isBand radiusBPmust cut the circlePCFinP[viii.,Cor.3]; but it touches it (hyp.) also inP, which is impossible.Hence the centre of the second circlemust be in the linePE.
Observation.—Propositionsxi,xii., may both be included in one enunciation as follows:—“If twocircles touch each other at any point, the centres and that point are collinear.” And thislatter Proposition is a limiting case of the theorem given in Propositioniii.,Cor.4,that “The line joining the centres of two intersecting circles bisects the common chordperpendicularly.”
Suppose the circle whose centre isOand one of the points of intersectionAto remain fixed,while the second circle turns round that point in such a manner that the second point ofintersectionBbecomes ultimately consecutive toA; then, since the lineOO′alwaysbisectsAB, we see that whenBultimately becomes consecutive toA, the lineOO′passesthroughA. In consequence of the motion, the common chord will become in the limita tangent to each circle, as in the second diagram.—Comberousse,Géométrie Plane,page57.
Cor.1.—If two circles touch each other, their point of contact is the union of two points ofintersection. Hence a contact counts for two intersections.
Cor.2.—If two circles touch each other at any point, they cannot have any other commonpoint. For, since two circles cannot have more than two points common [x.], and that the point ofcontact is equivalent to two common points, circles that touch cannot have any otherpoint common. The following is a formal proof of this Proposition:—LetO,O′be thecentres of the two circles,Athe point of contact, and letO′lie betweenOandA; take anyother pointBin the circumference ofO. JoinO′B; then [vii.]O′Bis greater thanO′A;therefore the pointCis outside the circumference of the smaller circle. HenceBcannot becommon to both circles. In like manner, they cannot have any other common point butA.
PROP.XIII.—Theorem.Two circles cannot have double contact, that it, cannot touch each other in twopoints.
PROP.XIII.—Theorem.Two circles cannot have double contact, that it, cannot touch each other in twopoints.
Dem.—1. If possible let two circles touch each other at two pointsAandB. Now since the two circles touch each other inA, the line joining their centres passes throughA[xi.]. In like manner, it passes throughB. Hence the centres and the pointsA,Bare in one right line; thereforeABis a diameter of each circle. Hence, ifABbe bisected inE,Emust be the centre of each circle—that is, the circles are concentric—which[v.]is impossible.
2. If two circles touched each other externally in two distinct points, then [xii.] the line joining the centres should pass through each point,which isimpossible.
Or thus:Draw a line bisectingABat right angles. Then this line [i.,Cor.1] must pass through the centre of each circle, and therefore [xi.xii.] must pass through each point of contact, which is impossible.Hence two circles cannot have doublecontact.
This Proposition is an immediate inference from the theorem [xii.,Cor.1], that a point ofcontact counts for two intersections, for then two contacts would be equivalent to four intersections;but there cannot be more than two intersections [x.]. It also follows from Prop.xii.,Cor.2, that iftwo circles touch each other in a pointA, they cannot have any other point common;hence theycannot touch again inB.
Exercises.
Exercises.
1.If a variable circle touch two fixed circles externally, the difference of the distances of itscentre from the centres of the fixed circles is equal to the difference or the sum of their radii,according as the contacts are of the same or of opposite species (Def.iv.).
2.If a variable circle be touched by one of two fixed circles internally, and touch the other fixedcircle either externally or internally, the sum of the distances of its centre from the centres of thefixed circles is equal to the sum or the difference of their radii, according as the contact with thesecond circle is of the first or second kind.
3.If through the point of contact of two touching circles any secant be drawn cutting the circlesagain in two points, the radii drawn to these points are parallel.
4.If two diameters of two touching circles be parallel, the lines from the point of contact to theextremities of one diameter pass through the extremities of the other.
PROP.XIV.—Theorem.In equal circles—1. equal chords(AB,CD)are equally distant from thecentre.2. chords which are equally distant from the centre are equal.
PROP.XIV.—Theorem.In equal circles—1. equal chords(AB,CD)are equally distant from thecentre.2. chords which are equally distant from the centre are equal.
Dem.—1. LetObe the centre. Draw the perpendicularsOE,OF. JoinAO,CO. Then becauseABis a chord in a circle, andOEis drawn from the centre cutting it at right angles, it bisects it [iii.]; thereforeAEis the half ofAB. In like manner,CFis the half ofCD; butABis equal toCD(hyp.). ThereforeAEis equal toCF[I., Axiomvii.]. And becauseEis a right angle,AO2is equal toAE2+EO2. In like manner,CO2is equal toCF2+FO2; butAO2is equal toCO2. ThereforeAE2+EO2is equal toCF2+FO2; andAE2has been proved equal toCF2. HenceEO2is equal toFO2; thereforeEOis equal toFO.HenceAB,CDare(Def.vi.)equally distant from thecentre.
2. LetEObe equal toFO, it is required to proveABequal toCD. The same construction being made, we have, as before,AE2+EO2equal toCF2+FO2; butEO2is equal toFO2(hyp.). HenceAE2is equal toCF2, andAEis equal toCF; butABis double ofAE, andCDdouble ofCF.ThereforeABis equal toCD.
Exercise.
Exercise.
If a chord of given length slide round a fixed circle—1. the locus of its middle point is a circle; 2.the locus of any point fixed in the chord is a circle.
PROP.XV.—Theorem.
PROP.XV.—Theorem.
The diameter(AB)is the greatest chord in a circle; and of the others, the chord(CD)which is nearer to the centre is greater than(EF)one more remote, and thegreater is nearer to the centre than the less.
Dem.—1. JoinOC,OD,OE, and draw the perpendicularsOG,OH; then becauseOis the centre,OAis equal toOC[I., Def.xxxii.], andOBis equal toOD. HenceABis equal to the sum ofOCandOD; but the sum ofOC,ODis greater thanCD[I.xx.].ThereforeABis greater thanCD.
2. Because the chordCDis nearer to the centre thanEF,OGis less thanOH; and since the trianglesOGC,OHEare right-angled, we haveOC2=OG2+GC2, andOE2=OH2+HE2; thereforeOG2+GC2=OH2+HE2; butOG2is less thanOH2; thereforeGC2is greater thanHE2, andGCis greater thanHE, butCDandEFare the doubles ofGCandHE.HenceCDis greater thanEF.
3. LetCDbe greater thanEF, it is required to prove thatOGis less thanOH.
As before, we haveOG2+GC2equal toOH2+HE2; butCG2is greater thanEH2; thereforeOG2is less thanOH2.HenceOGis less thanOH.
Exercises.
Exercises.
1.The shortest chord which can be drawn through a given point within a circle is theperpendicular to the diameter which passes through that point.
2.Through a given point, within or without a given circle, draw a chord of length equal to thatof a given chord.
3.Through one of the points of intersection of two circles draw a secant—1.the sum of whosesegments intercepted by the circles shall be a maximum; 2.which shall be of any length less thanthat of the maximum.
4.Three circles touch each other externallyatA,B,C; the chordsAB,ACof two of them areproduced to meet the third again in the pointsDandE; prove thatDEis a diameter of the thirdcircle, and parallel to the line joining the centres of the others.
PROP.XVI.—Theorem.
PROP.XVI.—Theorem.
1. The perpendicular(BI)to the diameter(AB)of a circle at its extremity(B)touches the circle at that point.2. Any other line(BH)through the same point cutsthe circle.
Dem.—1. Take any pointI, and join it to the centreC. Then because the angleCBIis a right angle,CI2is equal toCB2+BI2[I.xlvii.]; thereforeCI2is greater thanCB2. HenceCIis greater thanCB, and the pointI[note on I., Def.xxxii.] is without the circle. In like manner, every other point inBI, exceptB, is without the circle.Hence, sinceBImeets the circle atB, but does not cut it, it must touchit.
2. To prove thatBH, which is not perpendicular toAB, cuts the circle. DrawCGperpendicular toHB. NowBC2is equal toCG2+GB2. ThereforeBC2is greater thanCG2, andBCis greater thanCG. Hence [note on I., Def.xxxii.] the pointGmust be within the circle, and consequently the lineBGproduced must meet the circle again,and must therefore cut it.
This Proposition may be proved as follows:
At every point on a circle the tangent is perpendicular to the radius.
LetPandQbe two consecutive points on the circumference. JoinCP,CQ,PQ; producePQboth ways. Now sincePandQare consecutive points,PQis a tangent (Def.iii.). Again, the sum of the three angles of the triangleCPQis equal to two right angles; but the angleCis infinitely small, and the others are equal. Hence each of them is a right angle.Therefore the tangent is perpendicular to thediameter.
Or thus:A tangent is a limiting position of a secant, namely, when the secant moves out until the two points of intersection with the circle become consecutive; but the line through the centre which bisects the part of the secant within the circle [iii.] is perpendicular to it.Hence, in the limit the tangent is perpendicular to the line fromthe centre to the point of contact.
Or again:The angleCPRis always equal toCQS; hence, whenPandQcome together each is a right angle,and the tangent is perpendicular to theradius.
Exercises.
Exercises.
1.If two circles be concentric, all chords of the greater which touch the lesser areequal.
2.Draw a parallel to a given line to touch a given circle.
3.Draw a perpendicular to a given line to touch a given circle.
4.Describe a circle having its centre at a given point—1.and touching a given line; 2.andtouching a given circle. How many solutions of this case?
5.Describe a circle of given radius that shall touch two given lines. How many solutions?
6.Find the locus of the centres of a system of circles touching two given lines.
7.Describe a circle of given radius that shall touch a given circle and a given line, or that shalltouch two given circles.
PROP.XVII.—Problem.From a given point(P)without a given circle(BCD)to draw a tangent to thecircle.
PROP.XVII.—Problem.From a given point(P)without a given circle(BCD)to draw a tangent to thecircle.
Sol.—LetO(fig. 1) be the centre of the given circle. JoinOP, cutting the circumference inC. WithOas centre, andOPas radius, describe the circleAPE. ErectCAat right angles toOP. JoinOA, intersecting the circleBCDinB. JoinBP;it will be the tangent required.
Dem.—SinceOis the centre of the two circles, we haveOAequal toOP, andOCequal toOB. Hence the two trianglesAOC,POBhave the sidesOA,OCin one respectively equal to the sidesOP,OBin the other, and the contained angle common to both. Hence [I.iv.] the angleOCAis equal toOBP; butOCAis a right angle (const.); thereforeOBPis a right angle, and [xvi.]PBtouches the circle atB.
Cor.—IfAC(fig. 2) be produced toE,OEjoined, cutting the circleBCDinD, and the lineDPdrawn,DPwill be another tangent fromP.
Exercises.
Exercises.
1.The two tangentsPB,PD(fig.2) are equal to one another, because the square of each isequal to the square ofOPminus the square of the radius.
2.If two circles be concentric, all tangents to the inner from points on the outer areequal.
3.If a quadrilateral be circumscribed to a circle, the sum of one pair of opposite sides is equal tothe sum of the other pair.
4.If a parallelogram be circumscribed to a circle it must be a lozenge, and its diagonalsintersect in the centre.
5.IfBDbe joined, intersectingOPinF,OPis perpendicular toBD.
6.The locus of the intersection of two equal tangents to two circles is a right line (called theradical axisof the two circles).
7.Find a point such that tangents from it to three given circles shall be equal. (This point iscalled the radical centreof the three circles.)
8.The rectangleOF.OPis equal to the square of the radius.
Def.Two points, such asFandP, the rectangle of whose distancesOF,OPfrom thecentre is equal to the square of the radius, are called inverse pointswith respect to thecircle.
9.The intercept made on a variable tangent by two fixed tangents subtends a constant angle atthe centre.
10.Draw a common tangent to two circles. Hence, show how to draw a line cutting two circles,so that the intercepted chords shall be of given lengths.