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1 Label carefully each of the following: Circle Geometry labelling activity radius arc diameter centre chord sector major segment tangent circumference minor segment Board of Studies 1

2 These are the terms the Board of Studies says you should know for this topic: Circle Geometry Definitions of circle, centre, radius, diameter, arc, sector, segment, chord, tangent, concyclic points, cyclic quadrilateral, an angle subtended by an arc or chord at the centre and at the circumference, and of an arc subtended by an angle should be given. You should also know what a secant is! question: Do we know them all? 2

3 Assumption: Equal arcs on circles of equal radii subtend equal angles at the centre, and conversely. Circle Geometry (Conversely means that the opposite is also true, i.e. that equal angles at the centre subtend equal arcs on circles of equal radii!) While you do not HAVE to be able to prove this...you should be able to do so. question: How? 3

4 Theorem: proof not examined Circle Geometry Equal angles at the centre stand on equal chords and converse Almost the same as the last (assumption), chords drawn from A to B and D to J are equal. question: Can you construct the proof? 4

5 Theorem: proof not examined The angle at the centre is twice the angle at the circumference subtended by the same arc. Circle Geometry geogebra activity x y Pull Proof: <OAB = <OBA =x(base < AOB) Then <BOD = 2x (exterior angle AOB) <OAC = <OCA = y (base < AOC) Then <COD = 2y (exterior angle AOC) <BAC = x+y, <BOC = 2x+2y So angle at centre = twice angle at circumference x y 5

6 Theorem: proof not examined Two circles touch if they have a common tangent at the point of contact. Circle Geometry 6

7 Theorem: proof not examined The tangent to a circle is perpendicular to the radius drawn to the point of contact, and converse. Circle Geometry question: Why? geogebra activity 7

8 Theorem: proof IS examinable The perpendicular from the centre of a circle to a chord bisects the chord Circle Geometry geogebra activity Pull Proof: AO=BO (radii circle) AC = common side <OCA = <OCB (OC AB) So OCA OCB (RHS rule) AC=BC (corresp. sides, ) So OC bisects AB 8

9 Theorem: proof IS examinable The line from the centre of a circle to the midpoint of a chord is perpendicular to a chord Circle Geometry This follows logically from the previous theorem Pull geogebra activity Proof: OX=OY (radii of circle) OY=common side XY=ZY (Y midpoint of XZ) So by SSS rule, OXY OZY <OYX=<OYZ=90 0 (equal supp. angles) So OY is perpendicular to XZ 9

10 Theorem: proof IS examinable Equal chords in equal circles are equidistant from the centres, and converse. Circle Geometry geogebra activity Pull Proof: Let CD=AB <OEB=<OFD=90 0 (data) OB=OD (equal radii) BE= ½ AB (OE bisects AB) DF = ½ CD (OF bisects CD) BE= DF By RHS OEB OFD So OE=OF (corresp. sides congruent triangles) 10

11 Theorem: proof IS examinable Any 3 non-collinear points are concyclic. They lie on a unique circle, with centre at the point of intersection of the perpendicular bisectors of the intervals joining these points. Circle Geometry To see this concept clearly, experiment using the 'circle through three points' option in geogebra. B A Proof: The perpendicular line through the midpoint C B Pull of AB goes through the centre of the circle (see theorem slide 7). Similarly with BC and AC. Where the three perpendicular bisectors meet MUST be the centre of the circle. A O C 11

12 Theorem: proof IS examinable Angles in the same segment are equal Circle Geometry Pull Proof: Join A and D to centre O <AOD= 2 x <ABD ( < at centre = 2 x < at circ.) <AOD = 2 x <ACD <ABD = <ACD 12

13 Theorem: proof IS examinable The angle in a semicircle is a right angle Circle Geometry geogebra activity Pull Proof: Follows on from angle at centre = 2 x angle at circumference. If angle at centre = 180 0, angle at circumference must be

14 Theorem: proof IS examinable Opposite angles in a cyclic quadrilateral are supplementary Converse is also true, PLUS this is a test for the four points to be CONCYCLIC Circle Geometry Pull Proof: Join B and D to O geogebra activity Obtuse <DOB = 2 x <BAD (< at centre=2 x < at circ) Reflex <DOB = 2 x <BCD (< at centre=2 x < at circ) Above two angles add to (revolution) so <BAD + <BCD = (half of ) 14

15 Theorem: proof IS examinable The exterior angle at a vertex of a cyclic quadrilateral equals the interior opposite angle Circle Geometry geogebra activity Proof: Let <BAD = θ (opp < cyclic quad) A D Pull Then <BCD = θ (supp. <) So <DCE = θ (supp. <) And <DCE = <BAD B C E 15

16 Theorem: proof IS examinable If an interval subtends equal angles at two points on the same side of it then the endpoints of the interval and the two points are concyclic Circle Geometry geogebra activity 16

17 Axiom: A tangent to a circle is perpendicular to the radius at that point. Converse is also true. Circle Geometry geogebra activity 17

18 Theorem: proof IS examinable The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Circle Geometry Firstly: What does 'alternate segment' mean? DF is a tangent to the circle at C. CE is a chord dividing the circle into two segments, CEA and CEB. The segment CEA is said to be alternate to <ECF The segment CEB is said to be alternate to <ECD NOTE: If you can't see a strategy when you look at a problem, it's most likely this one! Hard to see and you need to practise spotting it. geogebra activity 18

19 The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Circle Geometry Pull Proof: Draw in diameter CF and join EF Let <ECB = x 0 <FCB = 90 0 (tangent meets radius at 90 0 ) <FEC = 90 0 (angle in a semicircle) <FCE = x 0 <FEC = x 0 (angle sum of triangle) <EFC = <EDC 19

20 Theorem: proof IS examinable Tangents to a circle from an external point are equal Circle Geometry A Pull Proof: OA=OB (radii) OC = common side geogebra activity <CAO = <CBO = 90 0 (tangent meets radii) So by RHS rule, AOC BOC and AC = BC (corresp sides, cong. triangles) O C B 20

21 Theorem: proof IS examinable The products of the intercepts of two intersecting chords are equal Circle Geometry geogebra activity Pull Proof: <AED = <CEB (vert opp <) <DAB=<DCB (< on chord DB) <ADC=<CBA (< on chord AC) So ADE CBE (AAA) And AE = DE EC EB therefore AE EB = DE EC AE EB = DE EC 21

22 Theorem: proof IS examinable The square of the length of the tangent from an external point is equal to the product of the intercepts of the secant passing through this point Circle Geometry Pull Proof: <QPR = <PSR <Q is common By angle sum, third < in each also equal PQR SPQ PQ = PR = QR QS SP PQ PQ = QR QS PQ PQ 2 = QR QS PQ 2 = QR QS 22

23 Theorem: proof IS examinable When circles touch, the line between the centres passes through the point of contact' Circle Geometry geogebra activity Pull Proof: AB is a tangent to circle centre O <OCB = 90 0 (tangent meets radius) AB is a tangent to circle centre P <PCB = 90 0 (tangent meets radius) <OCB + <PCB = = So OCP is a straight line 23

24 Groves 24

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53 End of topic: Summary including theorems and examples. Make sure you can use them AND construct the proofs where appropriate. You don't need to complete every question in this file. Make sure you have done a number from the last exercise though. You will probably need me to check them. Make some post-its of theorems and stick them on mirrors and walls! Circle Geometry 53

54 Attachments jw_angles_standing_on_arc.ggb jw_line_from _centre_to_midpoint_chord_perpendicular_to_chord.ggb jw_perp_bisector_chord_passes_through_centre.ggb 17a_CONSTRUCT_Tangent_Perpendic_to_Radius.ggb 5a_CONSTRUCT_Common_Chords.ggb 1_Chords_From_Centre.ggb jw_angle_in_semicircle.ggb jw_opp_angles_cyclic_quadrilateral.ggb 10_Ext_angle_cyclic_quad.ggb alternate segment circle theorem.ggb tangent meets radius.ggb 11_Tangents_from_external_point.ggb 14_Product_of_intercepts_of_chords.ggb 16_Tangent_squared_equals_secant_intercepts_products.ggb jw_centres_of_touching_circles.ggb

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