IX Class NCERT Mathematics 9. Circles Notes (PDF)

# IX CLASS-MATHEMATICS 9.CIRCLES ## Chapter 9 ### **9. Circles** 1. **Circle:** A circle is a collection of all the points in a plane that are at a fixed distance from a fixed point on the plane. The fixed point "O" is called the center of the circle, and the fixed distance OA is called the...

# IX CLASS-MATHEMATICS 9.CIRCLES ## Chapter 9 ### **9. Circles** 1. **Circle:** A circle is a collection of all the points in a plane that are at a fixed distance from a fixed point on the plane. The fixed point "O" is called the center of the circle, and the fixed distance OA is called the radius of the circle. 2. **Diameter:** A line segment joining any two points on the circle that passes through the center is called the diameter (AB). 3. **Chord:** A line segment joining any two points on the circle is called a chord (CD). 4. **Arc:** The part of the circle between any two points on it is called an arc. 5. **Semicircle:** If the end points of an arc become the end points of a diameter, then such an arc is called a semicircular arc or a semicircle. 6. **Minor Arc:** If the arc is smaller than a semicircle, then the arc is called a minor arc. 7. **Major Arc:** If the arc is longer than a semicircle, then the arc is called a major arc. 8. **Segment:** The region between the chord and the minor arc is called the minor segment, and the region between the chord and the major arc is called the major segment. 9. **Sector:** The area enclosed by an arc and the two radii joining the center to the end points of an arc is called a sector. One is a minor sector, and another is a major sector. ### **Angle Subtended by a Chord at a Point:** 1. ∠PRQ is called the angle subtended by the line segment PQ at the point R, and ∠POQ is the angle subtended by the chord PQ at the center O. 2. ∠PRQ, and ∠PSQ are respectively the angles subtended by PQ at points R and S on the major and minor arcs PQ. ### **Theorem 9.1:** Equal chords of a circle subtend equal angles at the center. **Proof:** AB and CD are two equal chords of a circle with center O. - OA=OC (Radii of a circle) - OB=OD (Radii of a circle) - AB=CD (Given) - Therefore, ΔAOB = ΔCOD (By SSS congruence rule). - ∠AOB=∠COD (By CPCT) (Corresponding parts of congruent triangles). ### **Theorem 9.2:** If the angles subtended by the chords of a circle at the center are equal, then the chords are equal. **Proof:** The angles subtended by the chords AB and CD of a circle at the center O are ∠AOB and ∠COD respectively. - Given ∠AOB=∠COD - OA=OC (Radii of a circle) - OB=OD (Radii of a circle) - ∠AOB=∠COD (Given) - Therefore, ΔAOB = ΔCOD (By SAS congruence rule) - AB=CD (By CPCT) ### **EXERCISE 9.1** 1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centers. **Sol:** C1, C2 are two congruent circles. - In ΔAOB and ∠ACO'D - OA=O'C (Radii of congruent circles) - OB=O'D (Radii of congruent circles) - AB=CD (Given) - Therefore, ΔAOB = ΔCOD (By SSS congruence rule). - ∠AOB=∠CO'D (By CPCT) ### **Perpendicular from the Centre to a Chord** ### **Theorem 9.3** The perpendicular from the centre of a circle to a chord bisects the chord. **Proof:** AB is a chord for the circle with centre O and OMIAB. - Join O to A and B. - In ΔAMO and ΔBMO - OA=OB (Radii of a circle) - OM=OM (Common) - ∠AMO=∠BMO=90° (OP⊥AB) - ΔAMO = ΔBMO (By RHS congruence rule). - AM=BM (By CPCT) ### **Theorem 9.4:** The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. **Proof:** Let AB be a chord of a circle with centre O and O is joined to the mid-point M of AB. Join OA and OB. - In ΔAMO, and ΔBMO - OA=OB (Radii of a circle) - OM=OM (Common) - AM=BM (M is the midpoint of AB) - ΔAMO = ΔBMO (By SSS congruence rule) - ∠AMO = ∠BMO (By CPCT) - But ∠AMO and ∠BMO are linear pair angles. - So, ∠AMO = ∠BMO=90° ### **Equal Chords and their Distances from the Centre** The length of the perpendicular from a point to a line is the distance of the line from the point. ### **Theorem 9.5:** Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres). **Proof:** PQ and RS are two equal chords of a circle with center O. - OL, and OM are perpendiculars to PQ and RS respectively. - We know that the perpendicular from the centre of a circle to a chord bisects the chord. - PL=QL=RS/2, and SM=RM=RS/2. - But PQ=RS (Given) - ⇒PL=RM→(1) - In ΔPOL and ΔROM - ∠OLP=∠OMR=90° - OP=OR (Radii of the same circle) - PL=RM (from(1)) - Therefore, ΔPOL = ΔROM (by RHS rule). - OL=OM(by CPCT) - Hence proved. ### **Theorem 9.6:** Chords equidistant from the centre of a circle are equal in length. **Example 1:** If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection, prove that the chords are equal. **Sol:** AB and CD are two chords of a circle, with centre O intersecting at a point E. - PQ is a diameter through E, such that ∠AEQ= ∠DEQ. - Draw OLIAB and OM⊥CD. - ∠LEO=∠MEO→(1) - In ΔOLE and ΔOME -∠LEO =∠MEO (from(1)) - ∠OLE = ∠OME (=90°) - EO=EO (Common) - ΔOLE = ΔOME (AAS congruency rule) - ⇒OL=OM (by CPCT) - ⇒ AB=CD. ### **EXERCISE 9.2:** 1. Two circles of radii 5 cm and 3 cm intersect at two points, and the distance between their centres is 4 cm. Find the length of the common chord. **Sol:** Let A,B are the centres of the circles. PQ is the common chord. - AP=5 cm; BP=3 cm. - AB=4 cm. - Let BR = x, then AR= 4-x. - We know that if two circles intersect each other at 2 points, then the line joining their centres is the perpendicular bisector to their common chord. - PQLAB. - APRB and APRA are right-angled triangles. - PR2=AP2-AR2. - PR2= 52-(4-x)2 - PR2 = 25-(16-x2+8x) - PR2= 9-x2+8x→(1) - PR2=AP2-BR2 - PR2= 32-x2 - PR2= 9-x2→(2) - From (1) and (2) - 9-x2+8x=9-x2 - 8x=0 ⇒ x=0 - Point B and R coincide. - PR=PB=3 cm - PQ=2×PR=2×3=6 cm - Length of the chord=6 cm. 2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord. **Sol:** AB,CD are two chords and AB=CD. - Let AB,CD intersect at E. - Now we prove AE=DE and CE=BE. - OP and OQ are perpendiculars to AB and CD from O. - We know that the perpendicular from the centre of a circle to a chord bisects the chord. - AP=PB=CQ=QD→(1) - In ΔOPE and ΔOQE - ∠OPE =∠OQE = 90° - OP=OQ (Distance from centre to equal chords) - OE=OE (common) - ΔOPE = ΔOQE( RHS congruence rule) - PE=QE (By CPCT)→(2) - Now AE=AP+PE=DQ+QE[From (1)and (2)] - .. AE=DE - Given AB=CD - AB-AE=CD-DE (: AE=DE) - .. BE=CE - Hence proved. 3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords. **Sol:** AB,CD are two chords and AB=CD. - Let AB,CD intersect at E. - Now we show that ∠OEA=∠OED. - Let OP and OQ are perpendiculars to AB and CD from O. - In ΔOPE and ΔOQE - ∠OPE =∠OQE = 90° - OP=OQ (Distance from centre to equal chords) - OE=OE (common) - ΔOPE = ΔOQE(RHS congruence rule) - ∠OEP =∠OEQ (By CPCT) - ∠OEA = ∠OFD. 4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB=CD (see Fig.). **Sol:** Let C1 and C2 are two concentric circles with centre O. - A line intersects C1 and C2 at B, C and A, D. - Let OELAD. - We know that the perpendicular from the centre of a circle to a chord bisects the chord. - In circle C2: AE=ED →(1) - In circle C1: BE=EC→ (2) - From (1)-(2) - AE-BE=ED-EC - AB=CD. 5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip? **Sol:** Join OR,OS,OM,RS,MS. OS intersect RM at P. - In ΔORS and ΔOMS - OR=OM (Radii) - RS=MS (Given) - OS=OS (Common) - ΔORS=ΔOMS (By SSS congruence rule) - ∠ROS=∠MOS(By CPCT) - ..∠ROP=∠MOP→(1) - In ΔROP, and ΔMOP - RO=MO (Radii) - ∠ROP=∠MOP(From (1)) - OP=OP (Common) - ΔROP = ΔMOP (By SAS congruence rule) - RP=PM and ∠RPO=∠MPO(By CPCT)→(2) - ∠RPO=∠MPO=90° - So, OP bisects the chord RM and OP⊥RM. - Let RP=PM=y; OP=x ⇒ PS=5-x. - In ΔOPR - x2+y2=52 (From Pythagoras theorem) - y2=25-x2 (3) - In ΔRPS - (5-x)2+y2=62 - y2=36-(5-x)2 - y2=36-(25-10x+x2) - y2=36-25+10x-x2 - y2=11+10x-x2→(4) - From (3)and (4) - 11+10x-x2=25-x2 - 10x=25-11=14 - x2=14/10=1.4 cm - From (3) - y2=25-x2=25-(1.4)2 = 25 - 1.96 = 23.04 - y=√23.04= 4.8 cm - The distance between Reshma and Mandip = 2y = 2 × 4.8 cm = 9.6 cm. 6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone. **Sol:** Ankur (A), Syed(S) and David(D) - Let AS=SD=DA=2x - Let OMLAD - ΔAMO = ΔDMO(By RHS congruence rule) - AM=MD (By CPCT) - ... AM=MD=x - From ΔAAMS - SM2=(2x)2-x2 = 4x2 – x2 = 3x2 - SM=√3x2 = √3x - OM=SM- OS = √3x – 20 - From ΔAMO - OM2+AM2=OA2 - OM2=OA2-AM2 - OM2=202-x2 = 400-x2 - (√3x-20)2=400-x2 - 3x2-40√3x+400=400-x2 - 3x2+x2=40√3x - 4x2=40√3x - x=10√√3 m - The length of the string of each phone = 2x=2×10√3 = 20√3 m. ### **Angle Subtended by an Arc of a Circle** 1. If two chords of a circle are equal, then their corresponding arcs are congruent and conversely, if two arcs are congruent, then their corresponding chords are equal. 2. Congruent arcs (or equal arcs) of a circle subtend equal angles at the centre. ### **Theorem 9.7:** The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. **Proof:** Let O be the centre of the circle. PQ is an arc subtending ∠POQ at the centre. Let A be a point on the remaining part of the circle. - Draw ray OA. - Angles opposite to equal sides in a triangle are equal. - In ΔAOP, OA=OP (radii of the same circle) - ⇒∠OAP = ∠OPA = x(say) - Similarly In ΔAOQ, OA=OQ - ⇒∠OAQ =∠OQA = y(say) - Let ∠POB=p and ∠QOB=q. - p=x+x (exterior angle is equal to sum of the opposite interior angles) - p=2x. - Similarly q=2y - p+q=2x+2y - p+q=2(x+y) - ∠POQ = 2∠PAQ - For the case (iii), where PQ is the major arc, Reflex of ∠POQ = 2 ∠ PAQ ### **Theorem 9.8:** Angles in the same segment of a circle are equal. **Example: Angle in a semicircle is a right angle.** **Sol:** PQ is a diameter, and 'O' is the centre of the circle. - .. ∠POQ=180° [Angle on a straight line] - ∠POQ = 2∠PAQ [Angle subtended by an arc at the centre is twice the angle subtended by it at any other point on the circle] - ∠POQ 180° - ..∠PAQ = = 90°. - 2 ### **Theorem 9.9:** If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle. **Proof:** AB is a line segment, which subtends equal angles at two points C and D. That is ∠ACB =∠ADB. - R.T.P: A, B, C and D are concyclic. - let us draw a circle through the points A, C and B. Suppose it does not pass through the point D. Then it will intersect AD or extended AD at a point, say E (or E'). - If points A, C, E and B lie on a circle, ∠ACB=∠AEB. - But it is given that ∠ACB=∠ADB - ..∠AEB=∠ADB - This is not possible unless E coincides with D. - Cyclic Quadrilateral: A quadrilateral ABCD is called cyclic if all the four vertices A, B, C, D of it lie on a circle. - **Theorem 9.10:** The sum of either pair of opposite angles of a cyclic quadrilateral is 180º. - If ABCD is a Cyclic quadrilateral, then ∠A+∠C=180°, and ∠B+∠D=180°. - **Theorem 9.11:** If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic. - If in quadrilateral ∠A+∠C=180° and ∠B+∠D=180°, then ABCD is a Cyclic quadrilateral. **Example 2:** In Fig. 9.19, AB is a diameter of the circle, CD is a chord equal to the radius of the circle. AC and BD when extended intersect at a point E. Prove that ∠AEB= 60°. **Solution:** Join OC, OD and BC. - In ΔODC, OC=OD=DC. - .. ΔODC is an equilateral. - ⇒∠COD=60° - ∠CBD=1/2 ∠COD (By Angle subtended theorem) - ∠CBD=1/2 × 60° = 30° - ∠CBE=30° - ∠ACB=90° (angle subtended by semi - circle is 90°) - ∠BCE=180°-∠ACB =180°-90°=90° - ∠CEB= 90°-∠CBE=90°-30°=60°, i.e.,∠AEB=60°. **Example 3:** In Fig 9.20, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD. **Solution:** ∠DAC=∠DBC=55° (Angles in the same segment). - ..∠DAB=∠DAC+∠BAC=55°+45°= 100° - But ∠DAB + ∠BCD= 180° (Opposite angles of a cyclic quadrilateral) - So, ∠BCD=180°-100°= 80°. **Example 4:** Two circles intersect at two points A and B. AD and AC are diameters to the two circles (see Fig. 9.21). Prove that B lies on the line segment DC. **Solution:** Join AB. - ∠ABD=90° (Angle in a semicircle). - ∠ABC= 90° (Angle in a semicircle). - So, ∠ABD + ∠ABC=90° + 90° = 180° - Therefore, DBC is a line. That is B lies on the line segment DC. **Example 5:** Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any quadrilateral is cyclic. **Solution:** In Fig. 9.22, ABCD is a quadrilateral in which the angle bisectors AH, BF, CF and DH of internal angles A, B, C and D respectively form a quadrilateral EFGH. - Now, ∠FEH = ∠AEB= 180° – ∠EAB –∠EBA - = 1/2 180° – (∠A + ∠B). - ∠FGH = ∠CGD = 180° - 1/2 (∠C + ∠D). - Therefore, ∠FEH+∠FGH = 180° - 1/2 (∠A + ∠B) + 180° - 1/2 (∠C + ∠D) - = 360° - 1/2 (∠A + ∠B + ∠C + ∠D) = 360° - 1/2 (360°) = 360° - 180°= 180° - In EFGH the pair of opposite angles are supplementary. - So, the quadrilateral EFGH is cyclic. ### **EXERCISE 9.3** 1. In Fig. 9.23, A,B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC. **Sol:** ∠AOC=∠AOB+∠BOC=60°+30°=90°. - The angle subtended by an arc at the centre, is double the angle subtended by it at any point on the remaining part of the circle. - 2×∠ADC=∠AOC. - 2×∠ADC=90° - ∠ADC= 90°/2 = 45° 2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc. **Sol:** O is the centre of the circle, and AB is a chord with length of radius. - OA=OB=AB (Radius) - ΔABO becomes an equilateral triangle. - .. ∠AOB=60° - Let C be a point on the major arc and D be a point on the minor arc. - The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. - 2 × ∠ACB = ∠AOB - 2 × ∠ACB = 60° - ∠ACB = 60°/2=30° - Since A, B,C and D lie on the same circle. ADBC is a cyclic quadrilateral. - ∠ACB +∠ADB=180° (In a cyclic quadrilateral opposite angles are supplementary) - 30°+∠ADB=180° - ∠ADB=180°-30°= 150°. - Required angles are 150°, 30°. 3. In Fig. 9.24, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR. **Sol:** We know that, the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. - Reflex ∠POR=2×∠PQR - Reflex ∠POR=2×100°= 200° . - ∠POR=360°-Reflex ∠POR=360°- 200° = 160° - In ΔPOR, OP=OR (Radii of the circle) - ∠OPR=∠ORP = x (Angles of opposite to equal sides are equal) - ∠OPR + ∠ORP + ∠POR = 180° (Angle sum property of triangle) - x+x+160°=180° - 2x= 180°-160° = 20° - x=20°/2 =10° - ⇒∠OPR=10° 4. In Fig. 9.25, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC. **Sol:** In ΔABC, the sum of all angles = 180º - ∠BAC + ∠ABC + ∠ACB = 180° - ∠BAC + 69° + 31° = 180° - ∠BAC + 100° = 180° - ∠BAC = 180° - 100° = 80° - We know that angles in the same segment of a circle are equal. - ∠BDC= ∠BAC - ∠BDC= 80° 5. In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC. **Sol:** ∠BEC + ∠CED = 180° ( Linear pair) - 130° + ∠CED = 180° - ∠CED = 180° - 130° = 50° - In ADEC, the sum of all angles = 180° - ∠CDE + ∠CED + ∠ECD = 180° - ∠CDE + 50° +20° = 180° - ∠CDE + 70° = 180° - ∠CDE = 180°-70° = 110° - ∠CDB = 110° - We know that angles in the same segment of a circle are equal. - ∠BAC = ∠CDB = 110°. 6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB=BC, find ∠ECD. **Sol:** We know that angles in the same segment of a circle are equal. - ∠CAD=∠CBD=70° - ∠BAC=∠BDC=30° - In ΔABC, AB=BC - ∠BAC=∠BCA = 30°( Angles opposite to equal sides are equal) - ∠BAD=∠BAC+∠CAD=30°+70°= 100° - ABCD is a cyclic quadrilateral - ∠BAD + ∠BCD = 180° - 100° + ∠BCD = 180° - ∠BCD = 180° – 100° = 80° - ∠ECD = ∠BCD - ∠BCA = 80° - 30° = 50°. 7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle. **Sol:** ABCD is a cyclic quadrilateral. - AC and BD are diameters. - We know that, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. - 2 × ∠ADC = ∠AOC - 2 × ∠ADC = 180° - ∠ADC =180°/2 = 90° - In ABCD one angle is 90°, and diagonals are equal. - So, ABCD is a rectangle. 8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic. **Sol:** ABCD is a trapezium with AB || DC and AD=BC. - Draw DELAB and CFIAB - In ΔADE and ABCF - AD=BC (Given) - ∠AED=∠BFC=90° - DE=CF (Distance between parallel lines) - ΔADE = ΔBCF (By RHS congruence rule) - ∠DAE=∠CBF (By CPCT) - ∠DAB=∠CBA→(1) - ∠ADC+∠DAB=180°(co- interior angles are supplementary) -∠ADC+∠CBA=180° - Opposite angles are supplementary. - Hence, ABCD is a cyclic quadrilateral. 9. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 9.27). Prove that ∠ACP = ∠ QCD. **Sol:** Join AP and DQ. - For chord AP - ∠ABP = ∠ACP( Angles in the same segment are equal)→(1) - For chord DQ - ∠QBD = ∠QCD ( Angles in the same segment are equal)→(2) - ∠ABP = ∠QBD (vertically opposite angles)→(3) - From (1), (2) and (3) - ∠ACP = ∠QCD 10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side. **Sol:** Two circles are drawn on the sides AC and BC of the ∆ABC. - The circles intersected at D. Join CD. - ∠ADC = ∠BDC = 90°(Angle in semi circle) - ∠ADC +∠BDC = 90° + 90° - ∠ADC + ∠BDC = 180° - AADB is a straight angle - So, D lies on AB. 11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD. **Sol:** Draw a circle with diameter AC. - ABCD is a quadrilateral. - ∠B + ∠D = 90° + 90° = 180° - In quadrilateral ABCD, opposite angles are supplementary. - ABCD is a cyclic quadrilateral. - CD is a chord. - ∠CAD = ∠ CBD( Angles in the same segment in a circle) 12. Prove that a cyclic parallelogram is a rectangle. **Sol:** Let ABCD is a cyclic parallelogram. - In cyclic quadrilateral opposite angles are supplementary. - ∠A + C = 180° - But ∠A = ∠C (In a parallelogram opposite angle are equal) - ..∠A = ∠C = 90° - ABCD is a parallelogram and one interior angle is 90° - So, ABCD is a rectangle.