1   Fill in the blanks$\\$ (i) The centre of a circle lies in __________ of the circle. (exterior/interior)$\\$ (ii) A point, whose distance from the centre of a circle is greater than its radius lies in __________ of the circle. (exterior/interior)$\\$ (iii) The longest chord of a circle is a __________ of the circle.$\\$ (iv) An arc is a __________ when its ends are the ends of a diameter.$\\$ (v) Segment of a circle is the region between an arc and __________ of the circle.$\\$ (vi) A circle divides the plane, on which it lies, in __________ parts.

Solution :

(i) The centre of a circle lies in $\underline{\text{interior}}$ of the circle.$\\$ (ii) A point, whose distance from the centre of a circle is greater than its radius lies in $\underline{\text{exterior}}$ of the circle.$\\$ (iii) The longest chord of a circle is a $\underline{\text{diameter}}$ of the circle.$\\$ (iv) An arc is a $\underline{\text{semi-circle}}$ when its ends are the ends of a diameter.$\\$ (v) Segment of a circle is the region between an arc and $\underline{\text{chord }} $of the circle.$\\$ (vi) A circle divides the plane, on which it lies, in $\underline{\text{three}}$ parts.

2   Write True or False: Give reasons for your answers.$\\$ (i) Line segment joining the centre to any point on the circle is a radius of the circle.$\\$ (ii) A circle has only finite number of equal chords.$\\$ (iii) If a circle is divided into three equal arcs, each is a major arc.$\\$ (iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.$\\$ (v) Sector is the region between the chord and its corresponding arc.$\\$ (vi) A circle is a plane figure.$\\$

Solution :

(i)True.$\\$ All the points on the circle are at equal distances from the centre of the circle, and this equal distance is called as radius of the circle.$\\$ (ii)False$\\$ There are infinite points on a circle. Therefore, we can draw infinite number of chords of given length. Hence, a circle has infinite number of equal chords.$\\$ (iii) False$\\$ Consider three arcs of same length as $AB, BC,$ and $CA$. It can be observed that for minor arc $BDC, CAB$ is a major arc. Therefore, $AB, BC$, and $CA$ are minor arcs of the circle.$\\$

(iv) True.$\\$ Let $AB$ be a chord which is twice as long as its radius. It can be observed that in this situation, our chord will be passing through the centre of the circle.Therefore, it will be the diameter of the circle.$\\$

(v) False.$\\$ Sector is the region between an arc and two radii joining the centre to the end points of the arc. For example, in the given figure, $OAB$ is the sector of the circle.$\\$

(vi) True.$\\$ A circle is a two-dimensional figure and it can also be referred to as a plane figure.

3   Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Solution :

A circle is a collection of points which are equidistant from a fixed point. This fixed point is called as the centre of the circle and this equal distance is called as radius of the circle. And thus, the shape of a circle depends on its radius. Therefore, it can be observed that if we try to superimpose two circles of equal radius, then both circles will cover each other. Therefore, two circles are congruent if they have equal radius. Consider two congruent circles having centre $O$ and $O'$ and two chords $AB$ and $CD$ of equal lengths.$\\$ In $\Delta AOB $ and $\Delta CO’D,$$\\$ $AB = CD$ (Chords of same length)$\\$ $OA = O'C$ (Radii of congruent circles)$\\$ $OB = O'D $(Radii of congruent circles)$\\$ $\therefore \Delta AOB \cong \Delta CO'D$ (SSS congruence rule)$\\$ $\Rightarrow \angle AOB = \angle CO'D$ (By $CPCT)$$\\$ Hence, equal chords of congruent circles subtend equal angles at their centres.

4   Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal

Solution :

Let us consider two congruent circles (circles of same radius) with centres as $O$ and $O'.$ In $\Delta AOB$ and $\Delta CO'D,$$\\$ $\angle AOB = \angle CO'D$ (Given)$\\$ $OA = O'C$ (Radii of congruent circles)$\\$ $OB = O'D$ (Radii of congruent circles)$\\$ $\therefore \Delta AOB \cong \Delta CO'D$ ($SSS$ congruence rule)$\\$ $ AB = CD $(By $CPCT)$$\\$ Hence, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

5   Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Solution :

Consider the following pair of circles.$\\$

The above circles do not intersect each other at any point. Therefore, they do not have any point in common.$\\$

The above circles touch each other only at one point $Y$. Therefore, there is $1$ point in common.$\\$

The above circles touch each other at $1$ point $X$ only. Therefore, the circles have $1$ point in common.$\\$

These circles intersect each other at two points $G$ and $H.$ Therefore, the circles have two points in common. It can be observed that there can be a maximum of $2$ points in common. Consider the situation in which two congruent circles are superimposed on each other. This situation can be referred to as if we are drawing the circle two times.

6   Suppose you are given a circle. Give a construction to find its centre.

Solution :

The below given steps will be followed to find the centre of the given circle.$\\$ $\text{Step 1:}$ Take the given circle.$\\$ $\text{Step 2:}$ Take any two different chords $AB$ and $CD$ of this circle and draw perpendicular bisectors of these chords.$\\$ $\text{Step 3:}$ Let these perpendicular bisectors meet at point $O$. $\\$Hence, $O $ is the centre of the given circle.

7   If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.

Solution :

Consider two circles centered at point $O$ and $O’$, intersecting each other at point $A$ and $B$ respectively.$\\$ Join $AB. AB$ is the chord of the circle centered at $O$. Therefore, perpendicular bisector of $AB$ will pass through $O$.$\\$ Again, $AB$ is also the chord of the circle centered at $O’$. Therefore, perpendicular bisector of $AB$ will also pass through $ O’$.$\\$ Clearly, the centres of these circles lie on the perpendicular bisector of the common chord.

8   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.

Solution :

Let the radius of the circle centered at $O$ and $O'$ be $5 cm$ and $3 cm$ respectively.$\\$ $OA = OB = 5 cm$ $\\$ $O'A = O'B = 3 cm$ $\\$ $OO'$ will be the perpendicular bisector of chord $AB$.$\\$ $\therefore AC = CB$ $\\$ It is given that, $OO' = 4 cm$ $\\$ Let $OC$ be $x$. Therefore, $O'C$ will be $4 - x.$ $\\$ In $\Delta OAC,$ $\\$ $OA ^2 = AC^ 2 + OC^ 2$ $\\$ $ \Rightarrow 5^ 2 = AC^ 2 + x^ 2$ $\\$ $ \Rightarrow 25 - x ^2 = AC^ 2$ ... (1)$\\$ In $\Delta O'AC,$ $\\$ $O'A^ 2 = AC^ 2 + O'C ^2$ $\\$ $ \Rightarrow 3 ^2 = AC ^2 + (4 - x)^ 2$ $\\$ $ \Rightarrow 9 = AC ^2 + 16 + x^ 2 - 8x$ $\\$ $ \Rightarrow AC ^2 = -x^ 2 - 7 + 8x$ ... (2)$\\$ From Equations (1) and (2), we obtain$\\$ $25 - x^ 2 = -x^ 2 - 7 + 8x$ $\\$ $8x = 32$ $\\$ $x = 4 $ $\\$ Therefore, the common chord will pass through the centre of the smaller circle i.e.,$ O'$ and hence, it will be the diameter of the smaller circle.

9   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.

Solution :

Let $PQ$ and $RS $ be two equal chords of a given circle and they are intersecting each other at point $T.$

Draw perpendiculars $OV$ and $OU$ on these chords.$\\$ In $\Delta OVT$ and $\Delta OUT,$ $\\$ $OV = OU$ (Equal chords of a circle are equidistant from the centre)$\\$ $\angle OVT = \angle OUT$ (Each $ 90^o)$ $\\$ $OT = OT $(Common)$\\$ $\therefore \Delta OVT \cong \Delta OUT (RHS$ congruence rule)$\\$ $\therefore VT = UT $(By $CPCT)$ ... (1)$\\$ It is given that,$\\$ $PQ = RS .....(2)$ $\\$ $ \Rightarrow \dfrac{1}{2}PQ=\dfrac{1}{2}RS $ $\\$ $ \Rightarrow PV=RU.....(3)$ $\\$ On adding Equations (1) and (3), we obtain$\\$ $PV + VT = RU + UT$ $\\$ $\Rightarrow PT = RT $ ... (4)$\\$ On subtracting Equation (4) from Equation (2), we obtain$\\$ $PQ - PT = RS - RT$ $\\$ $\Rightarrow QT = ST $ ... (5)$\\$ Equations (4) and (5) indicate that the corresponding segments of chords $PQ$ and $RS$ are congruent to each other.

10   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.

Solution :

Let $PQ $ and $RS$ are two equal chords of a given circle and they are intersecting each other at point $T$.

Draw perpendiculars $OV$ and $OU$ on these chords.$\\$ In $\Delta OVT$ and $\Delta OUT,$$\\$ $OV = OU$ (Equal chords of a circle are equidistant from the centre)$\\$ $\angle OVT = \angle OUT$ (Each $90^o)$ $\\$ $OT = OT$ (Common)$\\$ $\therefore OVT \cong \Delta OUT $(RHS congruence rule)$\\$ $\therefore \angle OTV = \angle OTU (By CPCT)$ $\\$ Therefore, it is proved that the line joining the point of intersection to the centre makes equal angles with the chords.

11   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 figure).

Solution :

Let us draw a perpendicular $OM$ on line $AD.$

It can be observed that $BC$ is the chord of the smaller circle and $AD$ is the chord of the bigger circle.$\\$ We know that perpendicular drawn from the centre of the circle bisects the chord.$\\$ $\therefore BM = MC ... (1)$ $\\$ And, $AM = MD ... (2)$ $\\$ On subtracting Equation (2) from (1), we obtain $\\$ $AM - BM = MD - MC$ $\\$ $\therefore AB = CD$

12   Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius $5 m$ 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 $6 m$ each, what is the distance between Reshma and Mandip?

Solution :

$ A R = A S = \dfrac{6}{2} = 3 m$ $\\$ $OR = OS = OM = 5 m. $(Radii of the circle) In $\Delta OAR,$$\\$ $OA^2 + AR^2 = OR^2$$\\$ $OA^2 +(3m)^2 =(5m)^2$ $\\$ $OA^2 =(25-9)m^2 =16m^2$ $\\$ $OA = 4 m$ $\\$ $ORSM$ will be a kite $(OR = OM$ and $RS = SM)$$\\$ . We know that the diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.$\\$ $\angle RCS$ will be of $90^o$ and $RC = CM$ $\\$ Area of $\Delta ORS= \dfrac{1}{2}OA*RS $ $\\$ $\dfrac{1}{2}*RC*OS =\dfrac{1}{2}*4*6$ $\\$ $RC*5=24$ $\\$ $RC = 4.8$ $\\$ $RM = 2RC = 2(4.8) = 9.6$ $\\$ Therefore, the distance between Reshma and Mandip is$ 9.6 m.$

Draw perpendiculars $OA$ and $OB$ on $RS$ and $SM$ respectively.

13   A circular park of radius $20 m$ 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.

Solution :

It is given that $AS = SD = DA$ $\\$ Therefore, $\Delta ASD$ is an equilateral triangle. $OA $(radius) = $20 m$ $\\$ Medians of equilateral triangle pass through the circumcentre $(O)$ of the equilateral triangle $ASD$. We also know that medians intersect each other in the ratio $2:1$. As $AB$ is the median of equilateral triangle $ASD$, we can write$\\$ $\implies \dfrac{OA}{ OB}=\dfrac{2}{ 1}$ $\\$ $\implies \dfrac{20m}{OB}=\dfrac{2}{1}$ $\\$ $\implies OB =(\dfrac{20}{2})=10m$ $\\$ $AB = OA + OB = (20 + 10) m = 30 m$ $\\$ In $\delta ABD,$ $\\$ $AD^2 = AB^2 + BD^2$ $\\$ $AD^2 = (30)^2 + (\dfrac{AD}{2})^2$ $\\$ $AD^2=900+\dfrac{1}{4}AD^2$ $\\$ $\dfrac{3}{4}AD^2=900 $ $\\$ $ AD^2 =1200 $ $\\$ $AD=20\sqrt{3}$ $\\$ Therefore, the length of the string of each phone will be $20\sqrt{ 3} m.$ $\\$

14   In the given figure, $A, B$ and $C$ are three points on a circle with centre $O$ such that $\angle BOC = 30^o$ and $\angle AOB = 60^o$. If $D$ is a point on the circle other than the arc $ABC,$ find $\angle ADC.$

Solution :

It can be observed that $\angle AOC = \angle AOB + \angle BOC$ $\\$ $= 60° + 30°$$\\$ $= 90°$$\\$ We know that angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.$\\$ $\angle ADC=\dfrac{1}{2}\angle AOC=\dfrac{1}{2}(90^o)=45^o$

15   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.

Solution :

In $\Delta OAB,$ $\\$ $AB = OA = OB = $radius$\\$ $\therefore \Delta OAB$ is an equilateral triangle.$\\$ Therefore, each interior angle of this triangle will be of $60°. $$\\$ $\angle AOB = 60^o$ $\\$ $\angle ACB=\dfrac{1}{2}\angle AOB=\dfrac{ 1}{2}(60^o)=30^o$ $\\$ In cyclic quadrilateral $ACBD,$$\\$ $\angle ACB + \angle ADB = 180°$ (Opposite angle in cyclic quadrilateral)$\\$ $\therefore \angle ADB = 180° - 30° = 150°$ $\\$ Therefore, angle subtended by this chord at a point on the major arc and the minor arc are$ 30°$ and $150° $ respectively.

16   In the given figure, $\angle PQR = 100°,$ where $P, Q$ and $R $ are points on a circle with centre $O.$ Find $\angle OPR.$

Solution :

Consider $PR$ as a chord of the circle.$\\$ Take any point $S$ on the major arc of the circle. $PQRS$ is a cyclic quadrilateral$\\$ $\angle PQR + \angle PSR = 180°$ (Opposite angles of a cyclic quadrilateral) $\\$ $\therefore \Delta PSR = 180° - 100° = 80°$ $\\$ 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.$\\$ $\therefore \Delta POR = 2\angle PSR = 2 (80°) = 160°$ $\\$ In $\delta POR,$ $\\$ $OP = OR $(Radii of the same circle)$\\$ $\therefore \angle OPR = \angle ORP$ (Angles opposite to equal sides of a triangle) $\\$ $\angle OPR + \angle ORP + \angle POR = 180°$ (Angle sum property of a triangle)$\\$ $ 2\angle OPR + 160° = 180°$ $\\$ $2\angle OPR = 180° - 160° = 20^o$ $\\$ $\angle OPR = 10°$

17   In fig. 10.38, $\angle ABC=69^o, \angle ACB=31^o,$ find $ \angle BDC$?

Solution :

$\angle BAC =\angle BDC$ (angles in the same segment of the circle) $\\$ In $\delta ABC,$$\\$ $\angle BAC + \angle ABC + \angle ACB =180°$(Agnle sum property of a $\Delta $) Or, $\\$ $\angle BAC+69°+31°=180°$ $\\$ Or, $\angle BAC=180°-100°$ Or, $\angle BAC=80°$ $\\$ Thus, $\angle BDC=80°$

18   In the given figure, $A, B, C$ and $D$ are four points on a circle. $AC$ and $BD$ intersect at a point $E$ such that $\angle BEC = 130° $ and $\angle ECD = 20°$. Find $\angle BAC.$

Solution :

In $\delta CDE,$ $\\$ $\angle CDE + \angle DCE = \angle CEB$ (Exterior angle)$\\$ $\therefore \angle CDE + 20° = 130°$ $\\$ $\therefore \angle CDE = 110°$ $\\$ However, $\angle BAC = \angle CDE$ (Angles in the same segment of a circle) $\therefore \angle BAC = 110°$

19   $ABCD$ is a cyclic quadrilateral whose diagonals intersect at a point $E$. If $\angle DBC = 70°, \angle BAC$ is $30°$, find $\angle BCD.$ Further, if $AB = BC,$ find $\angle ECD.$

Solution :

For chord $CD,$ $\\$ $\angle CBD = \angle CAD$ (Angles in the same segment)$\\$ $\angle CAD = 70°$ $\\$ $\angle BAD = \angle BAC + \angle CAD = 30° + 70° = 100°$ $\\$ $\angle BCD + \angle BAD = 180°$ $\\$ (Opposite angles of a cyclic quadrilateral) $\angle BCD + 100° = 180°$ $\\$ $\angle BCD = 80°$ $\\$ In $\Delta ABC,$ $\\$ $AB = BC$ (Given)$\\$ $\therefore \angle BCA = \angle CAB $(Angles opposite to equal sides of a triangle) $\\$ $\therefore \angle BCA = 30°$ $\\$ We have, $\angle BCD = 80°$ $\\$ $\therefore \angle BCA + \angle ACD = 80°$ $\\$ $30° + \angle ACD = 80°$ $\\$ $\therefore \angle ACD = 50°$ $\\$ $\therefore \angle ECD = 50°$

20   If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Solution :

Let $ABCD$ be a cyclic quadrilateral having diagonals $BD$ and $AC$, intersecting each other at point $O.$ $\\$ $\angle BAD =\dfrac{1}{2}\angle BOD=\dfrac{180^o}{2}=90^o$ (Consider $BD$ as a chord)$\\$ $\angle BCD + \angle BAD = 180°$ (Cyclic quadrilateral) $\\$ $\angle BCD = 180° - 90° = 90°$ $\\$ $\angle ADC =\dfrac{1}{2}\angle AOC =\dfrac{1}{2}(180^o)=90^o$(Considering $AC$ as a chord) $\\$ $\angle ADC + \angle ABC = 180°$ (Cyclic quadrilateral)$\\$ $90° + \angle ABC = 180°$ $\\$ $\angle ABC = 90°$ $\\$ Each interior angle of a cyclic quadrilateral is of $90°$. Hence, it is a rectangle.

21   If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Solution :

Consider a trapezium $ABCD$ with $AB || CD$ and $BC = AD.$ $\\$ Draw $AM \perp CD $ and $BN \perp CD.$ $\\$ In $\Delta AMD$ and $\Delta BNC,$ $\\$ $AD = BC$ (Given)$\\$ $\angle AMD = \angle BNC $ (By construction, each is $90°$)$\\$ $AM = BN $(Perpendicular distance between two parallel lines is same)$\\$ $\therefore \Delta AMD \cong \Delta BNC (RHS $congruence rule)$\\$ $\therefore \angle ADC = \angle BCD (CPCT) ... (1)$ $\\$ $\angle BAD$ and $\angle ADC$ are on the same side of transversal $AD.$ $\\$ $\angle BAD + \angle ADC = 180°$ ... (2)$\\$ $\angle BAD + \angle BCD = 180°$ [Using Equation (1)]$\\$ This equation shows that the opposite angles are supplementary. Therefore, $ABCD$ is a cyclic quadrilateral.

22   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 the given figure). Prove that $\angle ACP = \angle QCD.$

Solution :

Join chords $AP$ and $DQ.$ $\\$ For chord $AP,$ $\\$ $\angle PBA = \angle ACP$ (Angles in the same segment)$\\$ For chord $DQ,$ $\\$ $\angle DBQ = \angle QCD $ (Angles in the same segment) $\\$ $ABD $ and $PBQ$ are line segments intersecting at $B$ .$\\$ $\therefore \angle PBA = \angle DBQ$ (Vertically opposite angles)$\\$ From Equations (1), (2), and (3), we obtain$\\$ $\angle ACP = \angle QCD$

23   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.

Solution :

Consider a $\Delta ABC.$ $\\$ Two circles are drawn while taking $AB $ and $AC$ as the diameter. $\\$ Let they intersect each other at $D $ and let $D $ not lie on $BC.$ $\\$ Join $AD.$ $\\$ $\angle ADB = 90°$ (Angle subtended by semi-circle)$\\$ $\angle ADC = 90°$ (Angle subtended by semi-circle)$\\$ $ \angle BDC = \angle ADB + \angle ADC = 90° + 90° = 180°$ $\\$ Therefore, $BDC$ is a straight line and hence, our assumption was wrong. $\\$ Thus, Point $D$ lies on third side $BC$ of $\Delta ABC.$

24   $ABC$ and $ADC$ are two right triangles with common hypotenuse $AC.$ Prove that $\angle CAD = \angle CBD.$

Solution :

$90° + \angle ACD + \angle DAC = 180°$ $\\$ $\angle ACD + \angle DAC = 90°$ ... (2)$\\$ Adding Equations (1) and (2), we obtain$\\$ $\angle BCA + \angle CAB + \angle ACD + \angle DAC = 180°$ $\\$ $(\angle BCA + \angle ACD) + (\angle CAB + \angle DAC) = 180°$ $\\$ $\angle BCD + \angle DAB = 180°$ ... (3)$\\$ However, it is given that $\\$ $\angle B + \angle D = 90° + 90° = 180°$ $\\$ From Equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral $ABCD $ is $180°.$ $\\$ Therefore, it is a cyclic quadrilateral.$\\$ Consider chord $CD$ .$\\$ $\angle CAD = \angle CBD$ (Angles in the same segment)

In $\Delta ABC,$ $\\$ $\angle ABC + \angle BCA + \angle CAB = 180° $ (Angle sum property of a triangle) $\\$ $90° + \angle BCA + \angle CAB = 180°$ $\\$ $\angle BCA + \angle CAB = 90° ... (1)$ $\\$ In $\Delta ADC,$ $\\$ $\angle CDA + \angle ACD + \angle DAC = 180°$ (Angle sum property of a triangle)

25   Prove that a cyclic parallelogram is a rectangle.

Solution :

Let $ABCD$ be a cyclic parallelogram.$\\$ $\angle A + \angle C = 180°$ (Opposite angles of a cyclic quadrilateral) $\\$ We know that opposite angles of a parallelogram are equal. $\\$ $\angle A = \angle C$ and $\angle B = \angle D $ $\\$ From Equation (1), $\angle A + \angle C = 180°$ $\\$ $\angle A + \angle A = 180°$ $\\$ $2\angle A = 180°$ $\\$ $\angle A = 90°$ $\\$ Parallelogram $ABCD$ has one of its interior angles as $90°.$ Therefore, it is a rectangle.

26   Prove that line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Solution :

Let two circles having their centres as $O$ and $O’$ intersect each other at point $A$ and $B$ respectively. Let us join $O’.$

In $\Delta AOO'$ and $BOO’,$ $\\$ $OA = OB$ (Radius of circle 1)$\\$ $O’A = O’B$ (Radius of circle 2)$\\$ $OO’ = OO’ $(Common)$\\$ $\therefore \Delta AOO’ \cong \Delta BO O’$ (By $SSS$ congruence rule)$\\$ $\angle OAO’ = \angle OBO’$ (By $CPCT$)$\\$ Therefore, line of centres of two intersecting circles subtends equal angles at the two points of intersection.

27   Two chords $AB$ and $CD$ of lengths $5 cm 11cm$ respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between $AB$ and $CD$ is $6 cm,$ find the radius of the circle.

Solution :

Draw $OM \perp AB $ and $ ON \perp CD$ .Join $OB$ and $OD.$

$BM=\dfrac{AB}{2}=\dfrac{5}{2} $ (Perpendicular from the center bisects the chord)$\\$ $nD=\dfrac{CD}{2}=\dfrac{11}{2}$ $\\$ Let $ ON $ be $x$.Therefore ,$OM $ will be $6-x$ $\\$ In $ \Delta MOB,$ $\\$ $OM^2+MB^2=OB^2$ $\\$ $(6-x)^2+(\dfrac{5}{2})^2=OB^2$ $\\$ $36+x^2-12x+\dfrac{25}{4}=OB^2.....(1)$ $\\$ In $\Delta NOD,$ $\\$ $ON^2+ND^2=OD^2$ $\\$ $x^2+(\dfrac{11}{2})^2=OD^2$ $\\$ $x^2+\dfrac{121}{4}=OD^2......(2)$ $\\$ We have $OB=OD $ (Radii of the same circle )$\\$ Therefore ,from Equations (1) and (2),$\\$ $36+x^2-12x+\dfrac{25}{4}-\dfrac{121}{4}$ $\\$ $12 x=36+\dfrac{25}{4}-\dfrac{121}{4}$ $\\$ $=\dfrac{144+25-121}{4}=\dfrac{48}{4}=12$ $\\$ $x=1$ $\\$ From Equation(2),$ \\$ $(1)^2+\dfrac{121}{4}=OD^2$ $\\$ $OD^2=\dfrac{121 }{4}+1=\dfrac{125}{4}$ $\\$ $OD=\dfrac{5}{2}\sqrt{5}$ $\\$ Therefore , the radius of the circle is $ \dfrac{5}{2}\sqrt{5} cm=5.6 cm $(approx)

28   The lengths of two parallel chords of a circle are $6 cm$ and $8 cm.$ If the smaller chord is at distance $4 cm$ from the centre, what is the distance of the other chord from the centre?

Solution :

Let $AB$ and $CD$ be two parallel chords in a circle centered at $O.$ $\\$ Join $OB$ and $OD.$ Distance of smaller chord $AB$ from the centre of the circle = $4 cm$$\\$ $OM = 4 cm$$\\$ $MB= \dfrac{AB}{2} =\dfrac{6}{2}=3cm$ $\\$ In $\Delta OMB,$ $\\$ $OM^2 + MB^2 = OB^2$ $\\$ $ (4)^2 + (3)^2 = OB^2$ $\\$ $ 16 + 9 = OB^2$ $\\$ $ OB= \sqrt{25}$ $\\$ $OB = 5 cm$ $\\$ In $\Delta OND,$ $\\$ $OD = OB = 5 cm$ (Radii of the same circle) $\\$ $ND= \dfrac{CD}{2}=\dfrac{8}{2} =4cm$$\\$ $ON^2 + ND^2 = OD^2$ $\\$ $ON^2 + (4)^2 = (5)^2$ $\\$ $ON^2 =25-16=9$ $\\$ $ON = 3 cm$ $\\$ Therefore, the distance of the bigger chord from the centre is $3 cm.$

29   Let the vertex of an angle $ABC$ be located outside a circle and let the sides of the angle intersect equal chords $AD$ and $CE$ with the circle. Prove that $\angle ABC$ is equal to half the difference of the angles subtended by the chords $AC$ and $DE$ at the centre.

Solution :

In $\Delta AOD$ and $\Delta COE,$ $\\$ $OA = OC$ (Radii of the same circle) $\\$ $OD = OE$ (Radii of the same circle) $\\$ $AD = CE $(Given)$\\$ $\therefore \Delta AOD \cong \Delta COE $ ($SSS$ congruence rule)$\\$ $\angle OAD = \angle OCE$ (By $CPCT$).. (1)$\\$ $\angle ODA = \angle OEC$ (By $CPCT$ )... (2)$\\$ Also,$\\$ $\angle OAD = \angle ODA$ (As $OA = OD$) . ... (3)$\\$ From Equations (1), (2), and (3), we obtain $\\$ $\angle OAD = \angle OCE = \angle ODA = \angle OEC$ $\\$ Let $\angle OAD = \angle OCE = \angle ODA = \angle OEC = x$ $\\$ In $\Delta OAC,$ $\\$ $OA = OC$ $\\$ $\therefore \angle OCA = \angle OAC $(Let a)$\\$ In $\Delta ODE,$ $\\$ $OD = OE$ $\\$ $\angle OED = \angle ODE$ (Let y)$\\$ $ADEC$ is a cyclic quadrilateral.$\\$ $\therefore \angle CAD + \angle DEC = 180° $(Opposite angles are supplementary) $\\$ $x + a + x + y = 180°$ $\\$ $2x + a + y = 180°$ $\\$ $y = 180^o - 2x - a ... (4)$ $\\$ However, $\angle DOE = 180^o - 2y$ $\\$ And, $\angle AOC = 180^o - 2a$ $\\$ $\angle DOE - \angle AOC = 2a - 2y = 2a - 2 (180^o - 2x - a) = 4a + 4x - 360° ... (5)$ $\\$ $\angle BAC + \angle CAD = 180^o$ (Linear pair)$\\$ $\therefore \angle BAC = 180^o - \angle CAD = 180o - (a + x)$ $\\$ Similarly, $\angle ACB = 180^o - (a + x)$ $\\$ In $\Delta ABC,$ $\\$ $\angle ABC + \angle BAC + \angle ACB = 180^o$ (Angle sum property of a triangle)$\\$ $\angle ABC = 180^o - \angle BAC -\angle ACB$ $\\$ $= 180o - (180o - a - x) - (180^o - a -x)$ $\\$ $= 2a + 2x - 180^o$ $\\$ $= 12[4a+4x-360°]$ $\\$ $\angle ABC = \dfrac{1}{2} [\angle DOE - \angle AOC] $ [Using Equation (5)]

30   Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Solution :

Let $ABCD$ be a rhombus in which diagonals are intersecting at point $O$ and a circle is drawn while taking side $CD$ as its diameter. We know that a diameter subtends $90°$ on the arc.$\\$ $\therefore \Delta COD = 90°$ $\\$ Also, in rhombus, the diagonals intersect each other at $90°$.$\\$ $\angle AOB = \angle BOC = \angle COD = \angle DOA = 90°$$\\$ Clearly, point $O$ has to lie on the circle.

31   $ABCD$ is a parallelogram. The circle through $A, B $and $C$ intersect $CD$ (produced if necessary) at $E.$ Prove that $AE = AD.$

Solution :

It can be observed that $ABCE$ is a cyclic quadrilateral and in a cyclic quadrilateral, the sum of the opposite angles is $180°.$ $\\$ $\angle AEC + \angle CBA = 180°$ $\\$ $\angle AEC + \angle AED = 180°$ (Linear pair)$\\$ $\angle AED = \angle CBA ... (1)$ $\\$ For a parallelogram, opposite angles are equal.$\\$ $\angle ADE = \angle CBA ... (2)$ $\\$ From (1) and (2),$\\$ $\angle AED = \angle ADE$ $\\$ $AD = AE$ (sides opposite to equal Angles of a triangle are equal)

32   $AC$ and $BD$ are chords of a circle which bisect each other. Prove that $\\$ (i)$ AC $ and $BD$ are diameters;$\\$ (ii) $ABCD $ is a rectangle.

Solution :

Let two chords $AB$ and $CD$ are intersecting each other at point $O.$$\\$ In $\Delta AOB$ and $\Delta COD,$ $\\$ $OA = OC $(Given)$\\$ $OB = OD$ (Given)$\\$ $\angle AOB = \angle COD$ (Vertically opposite angles) $\\$ $\therefore \Delta AOB \cong \delta COD (SAS$ congruence rule) $\\$ $AB = CD$ (By $CPCT$)$\\$ Similarly, it can be proved that $\Delta AOD \cong \Delta COB$ $\\$

$\therefore AD = CB$ (By $CPCT$)$\\$ Since in quadrilateral $ACBD$, opposite sides are equal in length, $ACBD$ is a parallelogram.$\\$ We know that opposite angles of a parallelogram are equal. $\\$ $\therefore \angle A = \angle C$ $\\$ However, $\angle A + \angle C = 180° (ABCD $ is a cyclic quadrilateral) $\\$ $\angle A + \angle A = 180°$ $\\$ $2\angle A = 180°$ $\\$ $\therefore \angle A = 90°$ $\\$ As $ACBD$ is a parallelogram and one of its interior angles is $90°$, therefore, it is a rectangle. $\\$ $\angle A$ is the angle subtended by chord $BD$.$\\$ And as $\angle A = 90°,$ therefore, $BD$ should be the diameter of the circle.$\\$ Similarly, $AC$ is the diameter of the circle.

33   Bisectors of angles $A, B$ and $C$ of a triangle $ABC$ intersect its circumcircle at $D, E$ and $F$ respectively. Prove that the angles of the triangle $DEF $ are $90^o-\dfrac{1}{2}A, 90^o-\dfrac{1}{2}B,$ and $90^o-\dfrac{1}{2}C.$

Solution :

It is given that $BE$ is the bisector of $\angle B.$ $\\$ $\therefore \angle ABE = \dfrac{\angle B}{2}$ $\\$ However, $\angle ADE = \angle ABE$ (Angles in the same segment for chord $AE$) $\\$ $\therefore \angle ADE = \dfrac{\angle B}{2}$ $\\$ Similarly, $\angle ACF = \angle ADF = \dfrac{\angle C}{2}$ (Angle in the same segment for chord $AF$)$\\$ $\angle D = \angle ADE + \angle ADF$$\\$

$=\dfrac{\angle B}{2}+\dfrac{\angle C}{2}$ $\\$ $=\dfrac{1}{2}(\angle B+\angle C)$ $\\$ $=\dfrac{1}{2}(180^o-\angle A)$ $\\$ $=90^o -\dfrac{1}{2}\angle A$ $\\$ Similarly, it can be proved that $\\$ $\angle E=90^o-\dfrac{1}{2}\angle B$ $\\$ $\angle F=90^o-\dfrac{1}{2}\angle C$

34   Two congruent circles intersect each other at points $A$ and $B$. Through A any line segment $PAQ$ is drawn so that $P, Q$ lie on the two circles. Prove that $BP = BQ.$

Solution :

$AB$ is the common chord in both the congruent circles. $\\$ $\therefore \angle APB = \angle AQB$ $\\$ In $\Delta BPQ,$ $\\$ $\angle APB = \angle AQB$ $\\$ $\therefore BQ = BP$ (Sides opposite to equal angles of a triangle are equal)

35   In any triangle $ABC,$ if the angle bisector of $\angle A$ and perpendicular bisector of $BC$ intersect, prove that they intersect on the circum-circle of the triangle $ABC.$

Solution :

Let perpendicular bisector of side $BC$ and angle bisector of $\angle A$ meet at point $D$.$\\$ Let the perpendicular bisector of side $BC$ intersect it at $E.$ $\\$ Perpendicular bisector of side $BC$ will pass through circumcentre $O$ of the circle. $\\$ $\angle BOC$ and $\angle BAC$ are the angles subtended by arc $BC$ at the centre and a point $A$ on the remaining part of the circle respectively. We also 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.$\\$ $\angle BOC = 2\angle BAC = 2\angle A ... (1)$ $\\$ In $\Delta BOE $ and $\Delta COE,$ $\\$ $OE = OE$ (Common)$\\$ $OB = OC $(Radii of same circle)$\\$ $\angle OEB = \angle OEC$ (Each $90°$ as $OD \angle BC)$ $\\$ $\therefore \Delta BOE \cong \Delta COE (RHS$ congruence rule) $\\$ $\angle BOE = \angle COE $ (By $CPCT$) ... (2)$\\$ However, $\angle BOE + \angle COE = \angle BOC$ $\\$ $\therefore \angle BOE +\angle BOE = 2 \angle A$ [Using Equations (1) and (2)]$\\$ $2\angle BOE = 2\angle A \angle BOE = \angle A$ $\\$ $\angle BOE = \angle COE = \angle A$ $\\$ The perpendicular bisector of side $BC$ and angle bisector of $\angle A$ meet at point $D$.$\\$ $\therefore \angle BOD = \angle BOE = \angle A ... (3)$ $\\$ Since $AD$ is the bisector of angle $\angle A, $ $\\$ $\angle BAD = \dfrac{\angle A}{2}$ $\\$ $\therefore 2\angle BAD = \angle A ... (4)$ $\\$ From Equations (3) and (4), we obtain$\\$ $\angle BOD = 2\angle BAD$ $\\$ This can be possible only when point $BD$ will be a chord of the circle. $\\$ For this, the point $D$ lies on the circumcircle.$\\$ Therefore, the perpendicular bisector of side $BC$ and the angle bisector of $\angle A$ meet on the circumcircle of triangle $ABC$.