# Introduction Circle Some terms related with a circle

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2 (ii) Interior of a circle The collection of all points of the plane which lie inside the circle i.e. the collection of all points of the plane such that < r form the interior of the circle. The collection of all points of the plane which either lie on the circle or are inside the circle form the circular region. (iii) Exterior of a circle The collection of all points of the plane which lie outside the circle i.e. the collection of all points of the plane such that > r form the exterior of the circle. Interior r r 2. hord of a circle a line segment joining any two points of a circle is called a chord of the circle. In the adjoining figure, is a chord of the circle with centre. The distance is called the length of the chord. 3. iameter of a circle chord of a circle which passes through its centre is called a diameter of the circle. diameter of a circle is the longest chord of the circle and all diameters have equal length. In the adjoining figure, is a diameter of the circle with centre. ote that and are both radii of the circle, so = = r. It follows that = 2r = 2 radius. Thus: Length of a diameter = 2 radius. ote that there are infinitely many lines passing through the point (centre of circle), so a circle has infinitely many diameters. Exterior hord iameter 4. Secant of a circle line which meets a circle in two points is called a secant of the circle. In the adjoining figure, line is a secant of the circle with centre. Secant 5. rc of a circle rc (continuous) part of a circle is called an arc of the circle. The arc of a circle is denoted by the symbol. In the adjoining figure, denotes the arc of the circle with centre. rc of a circle is divided into following categories: (i) inor and major arc n arc less than one-half of the whole arc of a circle is called a minor arc of the circle and an arc greater than one-half of the whole arc is called a major arc of the circle. Here, is the minor arc and R is the major arc. inor arc Understanding ISE mathematics Ix R ajor arc

3 Semicircle (ii) Semicircle When is a diameter, then both arcs are equal and each is called a semicircle. Thus, one-half of the whole arc of a circle is called a semicircle. ircumference (iii) ircumference The whole arc of a circle is called the circumference of the circle. The length of the circumference of a circle is the length of the whole arc. However, in general, the term circumference of a circle refers to its length. 6. Sector of a circle The part of the plane region enclosed by an arc of a circle and its two bounding radii is called a sector of the circle. The part containing the minor arc is called minor sector and the part containing the major arc is called major sector. 7. Segment of a circle chord of a circle divides its circular region into two parts. Each part of the circular region is called a segment of the circle. The part of the circular region containing the minor arc is called a minor segment and the part containing the major arc is called a major segment. ajor sector inor sector inor segment ajor segment 8. ngle subtended by an arc The angle subtended by the two bounding radii of an arc of a circle at the centre of the circle is called the angle subtended by the arc. In the adjoining figure, is the angle subtended by the minor arc of a circle with centre and the reflex is the angle subtended by the major arc at. If R is any point on the major arc, then R is called the angle subtended by the minor arc at the point R. R 9. ngle subtended by a chord Let be a chord of a circle with centre and R, S be points on the minor and major arcs of the circle (as shown in the adjoining figure). Join and, then is the angle subtended by the chord at the centre. Join R, R, S and S, then R and S are the angles subtended by the chord at the points R and S respectively on the minor and major arcs. R S 10. oncentric circles Two or more circles are called concentric circles if and only if they have same centre but different radii. circle 1279

4 11. Equal (or congruent) circles Two or more circles are called equal (or congruent) circles if and only if they have same radius. In the adjoining figures, two circles with centres and have equal radius (= r), so, these are equal circles. r (a) r (b) 14.2 hord roperties of ircles Theorem 14.1 The straight line drawn from the centre of a circle to bisect a chord, which is not a diameter, is perpendicular to the chord. Given. chord of a circle with centre, and bisects the chord. To prove.. onstruction. Join and. roof. In s and 1. = 1. Radii of same circle. 2. = 2. is mid-point of. 3. = 3. ommon S.S.S. axiom of congruency. 5. Α = 5. c.p.c.t.. 6. Α + = is a st. line. 7. Α = 90 Hence..E.. 7. From 5 and 6. Theorem 14.2 (onverse of theorem 14.1) The perpendicular to a chord from the centre of the circle bisects the chord. Given. chord of a circle with centre, and is perpendicular to the chord. To prove. =. onstruction. Join and. roof. In s and 1. = 1. Radii of same circle. 2. = 2. Each = 90, since. 3. = 3. ommon R.H.S. axiom of congruency. 5. =.E.. 5. c.p.c.t Understanding ISE mathematics Ix

5 Theorem 14.3 Equal chords of a circle are equidistant from the centre. Given. and are chords of a circle with centre, and =. To prove. and are equidistant from i.e. if and, then =. onstruction. Join and. roof. 1. = erpendicular from centre bisects the chord (Theorem 14.2). 2. = Same as above. 3. = In s and 3. = (given). 4. = 4. From = 5. Each = 90, and. 6. = 6. Radii of same circle R.H.S. axiom of congruency. 8. =.E.. 8. c.p.c.t.. Theorem 14.4 (onverse of theorem 14.3) hords of a circle that are equidistant from the centre of the circle are equal. Given. and are chords of a circle with centre ;, and =. To prove. =. onstruction. Join and. roof. In s and 1. = 1. Given. 2. = 2. Radii of same circle. 3. = 3. Each = 90, and R.H.S. axiom of congruency. 5. = 5. c.p.c.t.. 6. = erpendicular from centre bisects the chord. 7. = Same as above = 1 2 =..E.. 8. =, from 5. circle 1281

6 Theorem 14.5 There is one and only one circle passing through three given noncollinear points. Given. Three non-collinear points, and. To prove. ne and only one circle can be drawn passing through the points, and. onstruction. Join and. raw perpendicular bisectors of line segments and, say and RS respectively. Let these bisectors meet at. Join, and. roof. R S 1. = 1. lies on the perpendicular bisector of and every point on the perpendicular bisector of a line segment is equidistant from its end points. 2. = 2. lies on the perpendicular bisector RS of and every point on the perpendicular bisector of a line segment is equidistant from its end points. 3. = = 3. From 1 and If a circle is drawn with as centre and radius =, it passes through the points and. 5. There is only one circle passing through the points, and. 4. From 3, is equidistant from points, and. 5. since two lines can intersect at only one point, the perpendicular bisectors of and meet at only one point. Hence, one and only one circle can be drawn passing through three given non-collinear points. orollary 1. In the plane of a circle, the perpendicular bisector of a chord of a circle passes through its centre. orollary 2. The perpendicular bisectors of two (non-parallel) chords of a circle intersect at the centre of the circle. orollary 3. s there is one and only one circle passing through three non-collinear points, two different circles can meet atmost in two different points. Illustrative Examples Example 1. chord of length 16 cm is drawn in a circle of diameter 20 cm. alculate its distance from the centre of the circle. Solution. Let be a chord of a circle with centre such that = 16 cm and radius of the circle = = 1. diameter = cm = 10 cm. 2 2 From, draw. Since, is bisected at (Theorem 14.2) = 1 = cm = 8 cm Understanding ISE mathematics Ix

7 From right angled, by ythagoras theorem, we get 2 = = = = = 36 = 6 cm. Hence, the given chord is at a distance of 6 cm from the centre of the circle. Example 2. The centre of a circle of radius 13 units is the point (3, 6). (7, 9) is a point inside the circle. is a chord of the circle such that =. alculate the length of. Solution. Given the centre of the circle is (3, 6) and (7, 9) is a point on the chord of the circle such that = i.e. is mid-point of the chord. Join. Since bisects the chord, (Theorem 14.1). Radius of the circle = = 13 units (given). (3, 6) = ( 7 3) 2 + ( 9 6) 2 istance formula = = = 5 units From right angled, by ythagoras theorem, we get 2 = = = = = 144 = 12 units. The length of chord = 2 = (2 12) units = 24 units. Example 3. In the figure given below, the diameter of a circle with centre is perpendicular to the chord. If = 8 cm and = 2 cm, find the radius of the circle. Solution. Since i.e., is bisected at (Theorem 14.2), = 1 = 1. 8 cm = 4 cm. 2 2 Let radius of circle = r cm, then = = (r 2). From right angled, by ythagoras theorem, we get 2 = r 2 = (r 2) 2 r 2 = 16 + r 2 4r + 4 4r = 20 r = 5. Radius of the circle = 5 cm. Example 4. Two chords, of lengths 24 cm, 10 cm respectively of a circle are parallel. If the chords lie on the same side of centre and the distance between them is 7 cm, find the length of a diameter of the circle. Solution. Let be the centre of the circle. raw and. = 1 = cm = 12 cm and = 1 = cm = 5 cm. Since, points, and are collinear. = distance between and = 7 cm (given). (7, 9) circle 1283

8 Let = x cm, then = + = (x + 7) cm. Let the radius of the circle be r cm. From right angled s and (by ythagoras theorem), we get 2 = and 2 = r 2 = (12) 2 + x 2 (i) and r 2 = (x + 7) 2 (ii) From (i) and (ii), we get (x + 7) 2 = (12) 2 + x x x + 49 = x 2 14x = 70 x = 5. From (i), r 2 = (12) = = 169 r = 13. The length of a diameter of the circle = 2r = 26 cm. Example 5. is an isosceles triangle inscribed in a circle. If = = 25 cm and = 14 cm, find the radius of the circle. Solution. Let be the centre of the circle. From, draw. Since is isosceles with =, is the mid-point of i.e. is perpendicular bisector of, therefore, lies on. = 1 = cm = 7 cm. 2 2 From right angled, by ythagoras theorem, we get 2 = 2 2 = (25) = = 576 = 24 cm. Let r cm be the radius of the circle, then = = (24 r) cm. From right angled, by ythagoras theorem, we get 2 = r 2 = (24 r) 2 r 2 = r 2 48r 48r = 625 r = = Hence, the radius of the circle = cm. Example 6. Two chords and of a circle are equal. rove that the centre of the circle lies on the bisector of. Given. and are chords of a circle with centre, and =. 1 To prove. lies on the bisector of i.e. =. onstruction. Join and. roof. In s and 284 Understanding ISE mathematics Ix 1. = 1. Radii of same circle. 2. = 2. Given.

9 3. = 3. ommon S.S.S. axiom of congruency. 5. =.E.. 5. c.p.c.t.. Example 7. In a circle of radius 5 cm, and are two chords such that = = 6 cm. Find the length of the chord. Solution. Let be the centre of the circle. Let the bisector of meet the chord at. In s and 1. = (by const.) 2. = (given) 3. = (common) = and Β = (S..S. axiom of congruency) ( c.p.c.t. ) ut + = 180 ( is a st. line) = 90. Therefore, is perpendicular bisector of, and hence it passes through the centre of the circle. Let = y cm and = x cm, then = = (5 x) cm [ radius = 5 cm] From right angled, y 2 + x 2 = 5 2 (i) From right angled, y 2 + (5 x) 2 = 6 2 i.e. y 2 + x 2 10 x = 11 (ii) Subtracting (ii) from (i), we get 10 x = x = 14 x = 7 5. Substituting this value of x in (i), we get y = 25 y2 = = 25 Length of chord = 2 = cm = 48 5 y 2 = cm = 9 6 cm. y = 24 5 Example 8. rove that the line joining mid-points of two parallel chords of a circle passes through the centre of the circle. Given., are two chords of a circle with centre., and are mid-points of and respectively. To prove. passes through. onstruction. Join, and through draw a straight line parallel to. roof m is mid-point of, and the line drawn from the centre of circle to bisect a chord is perpendicular to it. 2. = F E circle 1285

10 3. E = E, alternate angles are equal. 4. E = similarly, is mid-point of and same reasons as above. 5. E + E = dding 3 and is a straight line. Hence, passes through..e.. 6. Sum of adjacent angles is 180. Example 9. If two equal chords of a circle intersect, prove that their segments will be equal. Given. and are chords of a circle with centre. and intersect at and =. To prove. (i) = (ii) =. onstruction. raw,. Join. roof. 1. = = erpendicular from centre bisects the chord. 2. = = Same as above. 3. = and = In s and 4. = 3. = (given). 4. equal chords of a circle are equidistant from the centre. (Theorem 14.3) 5. = 5. Each = = 6. ommon R.H.S. axiom of congruency. 8. = 8. c.p.c.t = + =. 10. = =. Hence (i) = and (ii) =..E.. 9. From 3 and 8, adding. 10. From 3 and 8, subtracting. Example 10. f two unequal chords of a circle, prove that greater chord is nearer to the centre of the circle. Given. and are chords of a circle with centre. >, and,. To prove. <. onstruction. Join and Understanding ISE mathematics Ix

11 roof. 1. = erpendicular from centre bisects the chord. 2. = Same as above. 3. > 3. > (given) = In, = = In, = = = ( 2 2 ) < 0 2 < 2 <..E.. 6. =, radii of same circle. 7. From 3, > 2 > is +ve. Example 11. If two circles intersect in two points, then prove that the line through their centres is perpendicular bisector of the common chord. Given. Two circles with centres,, and intersecting at points, so that is their common chord. To prove. is perpendicular bisector of i.e. = and = 90. onstruction. Join,, and. roof. In s and 1. = 1. Radii of same circle. 2. = 2. Radii of same circle. 3. = 3. ommon S.S.S. axiom of congruency. 5. = In s and 5. c.p.c.t.. 6. = 6. From = 7. Radii of same circle. 8. = 8. ommon S..S. axiom of congruency. 10. = and = 10. c.p.c.t = is a st. line 12. = 90 Hence, is perpendicular bisector of..e From 10 and 11. circle 1287

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