Circle Class 10 Maharashtra 2023-24 PDF

2023-24 Class - X Board: Maharashtra CIRCLE CIRCLE: The set of all points equidistant from a fix...

2023-24 Class - X Board: Maharashtra CIRCLE CIRCLE: The set of all points equidistant from a fixed point in a plane is called circle. The fixed point is called the centre of the circle. In the above figure, point P is the centre of the circle. BASIC TERMS USED IN CIRCLE: 2.1 Radius: Distance between centre of a circle and any point on the circle is called the radius. Seg PQ, Seg PA and Seg PB are radii. 2.2 Chord: A segment whose end points lie on a circle is called the chord. Seg CD and Seg AB are chords. 2.3 Diameter: A chord which passes through the centre of the circle is called the diameter. Seg AB is a diameter. Length of the diameter is double the radius. 2.4 Tangent: A line in the plane of a circle which touches the circle exactly in only one point is called tangent of the circle. The point at which the tangent touches the circle is called the point of contact. Line XY is tangent to the circle and point R is the point of the contact. 2.5 Secant: A line which intersects the circle in two distinct points is called the secant. Line l is a secant. Basic concepts related to circles. (1) 'A Perpendicular segment drawn from the centre of a circle to the chord bisects the chord.' Given: (i) A circle with centre P. (ii) Seg PM ⊥ chord AB, A – M – B. Conclusion: AM = BM. (2) 'A segment joining centre of a circle and the midpoint of the chord is perpendicular to the chord.' Given: (i) A circle with centre A. (ii) Point M is midpoint of chord PQ. Conclusion: Seg AM ⊥ chord PQ. (3) 'In a circle (or in congruent circles), congruent chords subtend congruent angles at the centre.' Given: (i) A circle with centre P (ii) Chord AB ≅ chord CD Conclusion: ∠APB ≅ ∠CPD (4) 'In a circle (or in congruent circles) if two of more chords subtend congruent angles at the centre, then they are congruent.' Given: (i) A circle with centre O. (ii) ∠POQ ≅ ∠ROS Conclusion: Chord PQ ≅ chord RS (5) 'In a circle (or in congruent circles), congruent chords are equidistant from the centre.' Given: (i) A circle with centre P. (ii) Seg PM ⊥ chord AB, A – M – B (iii) Seg PN ⊥ chord CD, C – N – D (iv) Chord AB ≅ chord CD Conclusion: PM = PN (6) 'In a circle (or in congruent circles), chords which are equidistant from the centre are congruent.' Given: (i) A circle with centre O. (ii) Seg OM ⊥ chord PQ, P – M – Q (iii) Seg ON ⊥ chord RS, R – N – S (iv) OM = ON Conclusion: Chord PQ ≅ chord RS (7) Congruent circles: Two or more circles are said to be congruent circles if their radii are equal. In the above figure, Radius PQ ≅ Radius AB. ∴ Both circles are congruent. (8) Concentric circles: Two or more circles with same centre but different radii are called concentric circles. (9) Touching circles: Two circles in the same plane having only one point in common are called (10) Intersecting circles: Two circles which have exactly two points in common are called intersecting circles. (11) Circles passing through one point. Two points and Three points (i) We shall first consider one point. There are infinite number of circles passing through a point. (ii) Circles passing through two distinct points. There can be infinite number of circles passing through two distinct points. Note: Centres of all these circles will pass through a line containing perpendicular bisector of the segment joining these two points. (iii) Circles passing through three points. Here, arises to cases. Case (I): When three points are non-collinear. We can draw exactly one circle passing through three noncollinear points. Case (II): When three points are collinear. We can not draw any circle passing through three collinear points. (12) Tangent theorem: A tangent at any point of a circle is perpendicular to the radius, through the point of contact. In above figure, line PQ is a tangent at A and Seg AO is the radius through the point of contact A. ∴seg OA ⊥ line PQ. (13) Converse of tangent theorem: A line perpendicular to a radius of a circle at its outer end is a tangent to the circle. In adjoining figure, line l is perpendicular to radius OA at its outer end A. ∴ line l is a tangent. (14) Tangent segment Theorem: Statement: The lengths of the two tangent segments to a circle drawn from an external point are equal. Given: (i) A circle with centre A. (ii) D is a point in the exterior of the circle. (iii) Points P and Q are the points of contact of the two tangents from D to the circle. In △PAD and △QAD, Seg PA ≅ AQ... (radii of the same circle) Seg AD ≅ Seg AD... (Common side) ∠APD ≅ ∠AQD = 90°...... (by Tangent theorem) ∴ △PAD ≅ △QAD …... (by Hypotenuse side test) ∴ Seg DP ≅ Seg DQ …… (c. s. c. t.) Example: In the adjoining figure, the radius of a circle with centre C is 6 cm, Line AB is a tangent at A? Answer the following questions. (i) What is the measure of ∠CAB? Why? (ii) What is the distance of point C from line AB? Why? (iii) d(A, B) = 6 cm, find d(B, C). (iv) What the measure of ∠ABC? Why? Solution: (i) Radius CA ⊥ Line AB... (Tangent Theorem) ∴ m ∠CAB = 90° …… (i) (ii) d(C, A) = 6 cm …… (Radius of circle) ∴ Distance of point C from line AB is 6 cm. (iii) In △CAB, ∠CAB = 90°...[From (i)] 2 2 2 ∴ BC = AC + AB......... (By Pythagoras theorem) = 62 + 62 BC 2 = 36 + 36 = 72 ∴ BC = 6√2 cm ……. (Taking square roots) (iv) In △ABC, ∠A = 90°... [From (i)] AC = AB …. (Given) ∴ ∠ACB = ∠ABC ….. (ii) (converse of isosceles : triangle theorem) In △CAB, ABC + ACB + CAB + = 180...(Sum of all angles of a triangle is 180°) ABC + ABC + 90 = 180 …….. [From (i) and (ii)]  2ABC = 180 − 90  2ABC = 90 ABC = 45 THEOREM: If two circles are touching circles, then the common point lies on the line joining their centres. 1. Externally Touching Circles: In the adjoining figure, two circles with centres O and A are touching externally at point P. ∴O–P–A 2. Internally Touching Circles: In the adjoining figure, two circles with centres O and A are touching internally at point P. ∴O–A–P (1) Two circles having radii 3.5" " cm and 4.8" " cm touch each other internally. Find the distance between their centres. Given: (i) Two circles with centers A and B touch each other internally at point P. (ii) Radius of circle with centre A is 4.8 cm. (iii) Radius of circle with centre B is 3.5 cm. To Find: AB Solution: AP = 4.8 cm, BP = 3.5 cm …… (Given) A–B–P...(When two circles touch each other, the point of contact lies on the line joining the centres.)  AP = AB + BP 4.8 = AB + 3.5 AB = 4.8 − 3.5 ∴ AB = 1.3 cm AB = 1.3cm (2) Two circles having radii 5.5 cm, 4.2 cm touch each other externally. Find distance between their centres? Given: (i) Two circles with centres P and Q touch each other externally at point R. (ii) Radius of circle with centre P is 5.5 cm (iii) Radius of circle with centre Q is 4.2 cm To Find: PQ, QR  Solution: PR = 5.5 cm, QR = 4.2 cm …… (Given) P–R–Q (When two circles touch each other, the point of contact lies on the line joining the two centres.) PQ = PR + QR PQ = 5.5 + 4.2 PQ = 9.7cm  PQ = 9.7cm ARC OF A CIRCLE: A part of a circle is called an arc of a circle. In the adjoining figure, points A and B divide circle into two arcs, viz arc AXB and arc AYB. Measure of an arc: Measure of an arc equals to its corresponding central angle. m(arc PXQ) = m ∠POQ Types of arc: On the basis of measure of arc, arcs can be classified into three types: (1) Minor arc: An arc whose measure is less than 180° is called the minor arc. Note: Minor arcs are often named using two letters via in fig I and fig. II, arc AXB and arc PXQ can be named arc AB and arc PQ respectively. (2) Major arc: An arc measuring more than 180° is called the major arc. Measure of major arc = 360°, - Measure of minor arc. (3) Semicircle: An arc whose measure is 180° is called the semicircle. Note: A diameter divides a circle into two semicircles. Theorem: In a circle (or in congruent circles), congruent arcs have corresponding chords congruent. Given: (1) A circle with centre P. (2) arc AXB ≅ arc CYD To prove: chord AB ≅ chord CD...(i) [Given] Proof: arc AXB ≅ arc CYD m(arc AXB) = m ∠APB …..(i) [Definition of measure of a minor arc] m(arc CYD) = m ∠CPD …..(ii) ∴ m ∠APB = m ∠CPD …..(ii) [From (i), (ii) and (iii)] In △APB and △CPD, (i) seg PA ≅ seg PC (Radii of a circle), (ii) ∠APB ≅ ∠CPD [From (iv)] (iii) Seg PB ≅ Seg PD Radii of same circle ∴ △APB ≅ △CPD...(by SAS Test of congruence) ∴ chord AB ≅ chord CD (c...(c.c.t.)  Theorem: In a circle (or in congruent circles), congruent chords have their corresponding minon arcs congruent. Given: (i) A circle with centre P. (ii) chord DE ≅ chord FG To prove: arc DXE ≅ arc FYG Proof: (1) In △DPE and △FPG, (i) side DP ≅ side FP (ii) side PE ≅ side PG (Radii of the same circle) (iii) side DE ≅ side FG...(Given) (2) ∴ △DPE ≅ △FPG...(SSS Test) (3) ∴ ∠DPE ≅ ∠FPG ……(i) [c.a.c.t.] (4) m(arc DXE) = m ∠DPE ……(ii) [Definition of measure of a minor arc] (5) m(arc FYG) = m ∠FPG ……(iii) (6) ∴ m∠DPE = m∠FPG...[From (i), (ii) and (iii)] i.e. ∠DPE ≅ ∠FPG ∴ arc DXE ≅ arc FYG Arc addition Property: Inthead joining figure points A, X, B, Y and C are on the same circle and arc AXB and arc BYG has exactly one end point common. ∴ m(arc AXB) + m(arc BYC) = m(arc ABC) (2) In the adjoining figure, △QRS is an equilateral triangle. Prove: (1) arc RS ≅ arc QS ≅ arc QR (2) m(arc QRS) = 240°. Proof: △QRS is an equilateral triangle....(Given) ∴ chord QR ≅ chord RS ≅ chord QS...(Sides of equilateral Δ are equal) arc RS ≅ arc QS ≅ arc QR [In circle, congruent chords have corresponding minor arcs are congruent] Let m(arc RS) = m(arc QS) = m(arc QR) = x...(ii) [From (i) and supposition] ∴ m(arc RS) + m(arc QS) + m(arc QR) = 360°...(Measure of a circle is 360°) ∴ x + x + x = 360° ∴ 3x = 360° ∴ x = 120° ∴ m(arc RS) = m(arc QS) + m(arc QR) = 120°...(iii) ∴ m(arc QRS) = m(arc QR) + m(arc RS)...(Arc addition property) ∴ m(arc QRS) = 120° + 120° [From (iii)] ∴ m(arc QRS) = 240° Inscribed angle: An angle is said to be an inscribed angle, if (i) the vertex is on the circle (ii) both the arms are secants. In A the adjoining figure, ∠ABC is an inscribed angle, because vertex B lies on the circle and both the arms BA and BC are secants. In other words, ∠ABC is inscribed in arc ABC. Intercepted arc: Given an arc of the circle and an angle, if each side of the angle contains an end point of the arc and all other points of the arc except the end points lie in the interior of the angle, then the arc is said to be intercepted by the angle. (i) In figures (a), (b) and (c), ∠ABC has its vertex B outside the circle and intercepts two arcs. (ii) In figures (d) and (e), ∠ABC has its vertex on the circle and intercepts only one arc. (iii) In figure (f), ∠ABC has its vertex B inside the circle and intercepts only one arc. Inscribed Angle Theorem The measure of an inscribed angle is half of the measure of its intercepted arc. In the adjoining figure, ∠ABC is an inscribed angle and arc AXC is the intercepted arc. 1 mABC = m(arc AXC ) 2 ⇒ Corollary - 1: An angle inscribed in a semicircle is a right angle. In the adjoining figure, ∠ABC is inscribed in the semicircle ABC.  mABC = 90 ⇒ Corollary -2: Angles inscribed in the same arc are congruent. In the adjoining figure, ∠ABC and ∠ADC, both are inscribed in the same arc ABC.  ABC  ADC Note Inscribed angles, ∠ABC and ∠ADC both intercept the same arc AC. ∴ ∠ABC = ∠ADC Cyclic Quadrilateral: A quadrilateral whose all four vertices lie on a circle is called cyclic quadrilateral. In the adjoining figure, ABCD is cyclic, as all the four vertices A, B, C and D, lie on a circle. Theorem Statement: The opposite angles of a cyclic quadrilateral are supplementary. Given: ABCD is a cyclic quadrilateral. To Prove: m∠ABC + m∠ADC = 180° m∠BAD + m∠BCD = 180°  Proof: m∠ABC = 1/2 m(arc ADC) ………(i) (Inscribed angle theorem) m∠ADC = 1/2 m(arc ABC) ………(ii) Adding (i) and (ii), we get m∠ABC + m∠ADC 1 1 = m(arc ADC ) + m(arc ABC ) 2 2 mABC + mADC 1 = [m(arc ADC ) + m(arc ABC )] 2 1 mABC + mADC =  360... (Measure of a circle is 360°) 2 ∴ m∠ABC + m∠ADC = 180° …….. (iii) In ABCD, m∠BAD + m∠BCD + m∠ABC + m∠ADC = 360° (Sum of measure of angles of a quadrilateral is 360°) ∴ m∠BAD + m∠BCD + 180° = 360° …. [From (iii)] ∴ m∠BAD + m∠BCD = 180° Converse of cyclic quadrilateral theorem: If the opposite angles of a quadrilateral are supplementary, then it is a cyclic quadrilateral. In ABCD, if m∠A + m∠C = 180° or m∠B + m∠D = 180° then, ABCD is a cyclic quadrilateral. Corollary: An exterior angle of cyclic quadrilateral is congruent to the angle opposite to adjacent interior angle. In the adjoining figure, ABCD is cyclic ∠DCE is an exterior angle. ∴ ∠DCE ≅ ∠BAD If end points of a segment forms congruent angles on the same side of the segment, then the vertices of those angles and end points of the segment are concyclic. ∠ABD ≅ ∠ACD ∴ Points A, B, C and D are concyclic. Note: Concyclic means points lying on the same circle. TWO SPECIAL PROPERTIES: 1 mAPD = [m(arc AXD) + m(arc BYC )] 2 1 mPQR = [m(arc PXR) − m(arc AB)] 2 Example: In the adjoining figure, PQRS is a cyclic quadrilateral, side PQ ≅ side RQ. ∠PSR = 110°. Find (i) ∠PQR (ii) m (arc PQR) (iii) m (arc QR) (iv) ∠PRQ Solution: PQRS is a cyclic quadrilateral ….(Given) PQR + PSR = 180...(Cyclic quadrilateral theorem) PQR + 110 = 180 PQR = 180 − 110 ….. (i) mPQR = 70 1 PSR = m(arc PQR ). ……. (Inscribed angle theorem) 2 1 110 = m(arc PQR) …… (v) [From (iv)] 2 m(arc PQR) = 220 …….. (ii) ∴ In △PQR, side PQ ≅ side RQ …….. (Given) ∴ ∠QPR ≅ ∠QPR …….. (iii) (Isosceles triangle theorem)  In △PQR, ∠PQR + ∠PRQ + ∠QPR = 180° ….. (Sum of all angles of a triangle is 180°) ∴ 70° + ∠QPR + ∠QPR = 180° …(From (i) and (ii)] ∴ 2∠QPR = 180° – 70° ∴ 2∠QPR = 110° ∴ ∠QPR = 55° …… (iv) 1 QPR = m(arc QR )..(Inscribed angle theorem) 2 1  55 =  m(arc QR ) ….. [From (iv)] 2 m(arc QR) = 110 ……[From (iii) and (iv)] Example: In the adjoining figure, point O is the centre of the circle. Length of chord AB is equal to the radius of the circle. Find. (i) ∠AOB (ii) ∠ACB (iii) m(arc AB) ( iv) m(arc ACB) Solution: OA = OB...(i) (Radii of the same circle) AB = Radius of the given circle...(ii) (Given) ∴ AB = OA = OB …..[From (i) and (ii)] ∴ △OAB is an equilateral triangle...(By definition) ∴ ∠AOB = 60 …. (iii) (Measure of an angle of an equilateral triangle) m(arc AB) = m∠AOB...(Definition of measure of minor arc) ∴ m(arc AB) = 60° …. (iv) [From (iii)] 1 ACB = (arc AB ) ….... (Inscribed angle theorem ) 2 1 ACB =  60 = 30 ……. (v) [From (iv)] 2 m(arc ACB) + m(arc AB) = 360° ….... (Measure of a circle is 360°) m(arc ACB) + 60° = 360°...[From (v)] m(arc ACB) = 360° – 60° ∴ m(arc ACB) = 300° TANGENT SECANT THEOREM: If an angle with its vertex on the circle whose one side touches the circle and the other intersects the circle in two points, then the measure of the angle is half the measure of its intercepted arc. In the above three figures, ∠ABC has its vertex B on the circle, line BC is tangent to circle at B and ray BA is a secant. 1 mABC = m(arc ADB) 2 Segment of a circle: A secant divides the circular region into two parts. Each part is called a segment of the circle. Alternate segment: Each of the two segments formed by the secant of a circle is called alternate segment in relation with the other. Angle formed in a segment: An angle inscribed in the arc of a segment is called an angle formed in that segment. In the adjoining figure, secant AB divides the circular region into two segments R1 and R2. R1 and R2 are alternate segments in relation with each other. ∠ACB is inscribed in arc ACB of segment R1. ∴ ∠ACB is an angle formed in segment R1 Angles in Alternate Segment: If a line touches a circle and from the point of contact a chord is drawn, then the angles which this chord makes with the given line is equal respectively to the angle formed in the corresponding alternate segment. In the above figure, 1 mABC = m(arc AXB) ……. (i)...(Tangent secant theorem) 2 1 mADB = m(arc AXB)...(Inscribed angle theorem) 2 mABC = mADB …[ From (i) and (ii)] ABC  ADB  Theorem: If two secants of a circle intersect inside or outside the circle then the area of the rectangle formed by the two line segments corresponding to one secant is equal in area to the rectangle formed by the two line segments corresponding to the other. In the adjoining figure, chords AB and CD intersect each other at point O inside the circle. ∴ OA × OB = OC × OD In the adjoining figure, chords AB and CD intersect each other at point P outside the circle. ∴ OA × OB = OC × OD Tangent Secant Segment Theorem Statement: If a secant and a tangent of a circle intersect in a point outside the circle, then the area of the rectangle formed by the two-line segments corresponding to the secant is equal to the area of the square formed by the line segment corresponding to the other tangent. Given: (i) line PAB is a secant intersecting the circle at points A and B. (ii) line PT is a tangent to the circle at point T. To Prove: PA × PB = PT 2 Construction: Draw seg BT and seg AT. Proof: In △PTA and △PBT, ∠TPA ≅ ∠BPT (Common angle) ∠PTA ≅ ∠PBT (Angles in alternate segment) ∴ △PTA ∼ △PBT (By AA test of similarity) PT PA ∴ = (Corresponding sides of similar triangles) PB PT ∴ PA × PB = PT 2 Example: In the adjoining figure, m(arc WY) = 44, m(arc ZX) = 68 then, (i) Find m∠ZTX and m(arc WZ) (ii) If l(WT) = 4.8, l(TX) = 8, l(YT) = 6.4, then find l(TZ) (iii) If l(WX) = 12.8, l(YT) = 6, l(TX) = 6.4, then find l(WT) Solution: 1 (i) mZTX = [m(arc WY ) + m(arc ZX )] 2 1 = [44 + 68] 2 1 =  112 2 mZTX = 56 (ii) WT  TX = TZ YT...(Theorem on internal division of chords)  4.8  8 = TZ  6.4 4.8  8  TZ = 6.4  TZ = 6 units (iii) WX = WT + TX ……. (W – T – X)  12.8 = WT + 6.4  WT = 12.8 − 6.4  WT = 6.4 units Example: In the adjoining figure, (i) It m(arc CE) = 54°, m(arc BD) = 23°, then find ∠CAE (ii) If AB = 4.2, BC = 5.4, AE = 12, then find AD (iii) If AB = 3.6, AC = 9.0, AD = 5.4, then find AE. Solution: 1 (i) CAE = [m(arc CE ) − m(arc BD)] 2 1 = (54 − 23) 2 1 =  31 2 CAE = 15.5 (ii) AD  AE = AB  AC AC = AB + BC...(Theorem of external division of chords)  AD12 = 4.2  9.6 = 4.2 + 5.4 4.2  9.6  AD = = 9.6 12  AD = 3.36 units (iii) AE  AD = AC  AB...(Theorem of external division of chords)  AE  5.4 = 9  3.6 9  3.6  AE = 5.4  AE = 6 units Example: In the adjoining figure, point Q is the point of contact. If PQ = 12, PR = 8, then find PS and RS. Solution: Tangent PQ and secant PRS intersect each other at point P.  PQ2 = PR × PS...(Tangent secant segments theorem)  122 = 8  PS 12 12  = PS 8  PS = 18 units PS = PR + RS …… (P – R – S)  18 = 8 + RS  RS = 18 − 8  RS = 10 units Example: In the adjoining figure, point B is the point of contact and point O is the centre of the circle. Seg OE ⊥ Seg AD, if AB = 12, AC = 8, then find (i) AD (ii) DC (iii) DE Solution: Tangent AB and Secant ACD intersect at point A.  AB 2 = AC × AD...(Tangent secant segments theorem)  122 = 8  AD 12  12  = AD 8  AD = 18 units  AD = AC + DC …… (A – C – D)  18 = 8 + DC  DC = 18 − 8  DC = 10 units  Seg DE ⊥ chord CD ….. (Given) 1  DE =  CD (Perpendicular drawn from the centre of the circle to the chord bisects the chord) 2 1  DE = 10 [From (i)] 2  DE = 5 units PW Web/App - https://smart.link/7wwosivoicgd4 Library- https://smart.link/sdfez8ejd80if