Chapter 9: Circles
Exercise 9.1
Q1
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. 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 \( \Delta CO'D \):
\[ AB = CD \quad \text{(Chords of same length)} \]
\[ OA = O'C \quad \text{(Radii of congruent circles)} \]
\[ OB = O'D \quad \text{(Radii of congruent circles)} \]
\[ \therefore \Delta AOB \cong \Delta CO'D \quad \text{(SSS congruence rule)} \]
\[ \Rightarrow \angle AOB = \angle CO'D \quad \text{(By CPCT)} \]
Hence, equal chords of congruent circles subtend equal angles at their centres.
Q2
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 \( \Delta CO'D \):
\[ \angle AOB = \angle CO'D \quad \text{(Given)} \]
\[ OA = O'C \quad \text{(Radii of congruent circles)} \]
\[ OB = O'D \quad \text{(Radii of congruent circles)} \]
\[ \therefore \Delta AOB \cong \Delta CO'D \quad \text{(SAS congruence rule)} \]
\[ \Rightarrow AB = CD \quad \text{(By CPCT)} \]
Hence, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Exercise 9.2
Q1
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.
\[ OA = OB = 5 \text{ cm} \]
\[ O'A = O'B = 3 \text{ cm} \]
\( OO' \) will be the perpendicular bisector of chord AB.
\[ \therefore AC = CB \]
It is given that \( OO' = 4 \text{ cm} \). Let OC be \( x \). Therefore, \( O'C \) will be \( 4-x \).
In \( \Delta OAC \):
\[ OA^2 = AC^2 + OC^2 \]
\[ 5^2 = AC^2 + x^2 \Rightarrow 25 - x^2 = AC^2 \quad \dots(1) \]
In \( \Delta O'AC \):
\[ O'A^2 = AC^2 + O'C^2 \]
\[ 3^2 = AC^2 + (4-x)^2 \]
\[ 9 = AC^2 + 16 + x^2 - 8x \]
\[ AC^2 = -x^2 - 7 + 8x \quad \dots(2) \]
From equations (1) and (2):
\[ 25 - x^2 = -x^2 - 7 + 8x \]
\[ 8x = 32 \Rightarrow x = 4 \]
Since \( x = 4 \), the point C coincides with the centre \( O' \). Therefore, the common chord will pass through the centre of the smaller circle and will be its diameter.
\[ AC^2 = 25 - 4^2 = 25 - 16 = 9 \]
\[ AC = 3 \text{ cm} \]
Length of common chord \( AB = 2 AC = 6 \text{ cm} \).
Q2
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 \( \Delta OUT \):
\[ OV = OU \quad \text{(Equal chords are equidistant from centre)} \]
\[ \angle OVT = \angle OUT \quad \text{(Each } 90^{\circ} \text{)} \]
\[ OT = OT \quad \text{(Common)} \]
\[ \therefore \Delta OVT \cong \Delta OUT \quad \text{(RHS congruence rule)} \]
\[ VT = UT \quad \text{(By CPCT) ... (1)} \]
It is given that \( PQ = RS \). Therefore:
\[ \frac{1}{2} PQ = \frac{1}{2} RS \Rightarrow PV = RU \quad \dots(2) \]
On adding equations (1) and (2):
\[ PV + VT = RU + UT \Rightarrow PT = RT \quad \dots(3) \]
On subtracting equation (3) from \( PQ = RS \):
\[ PQ - PT = RS - RT \Rightarrow QT = ST \quad \dots(4) \]
Equations (3) and (4) show that the corresponding segments are equal.
Q3
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 be two equal chords intersecting at T. Draw perpendiculars OV and OU on these chords.
In \( \Delta OVT \) and \( \Delta OUT \):
\[ OV = OU \quad \text{(Equal chords equidistant from centre)} \]
\[ \angle OVT = \angle OUT \quad \text{(Each } 90^{\circ} \text{)} \]
\[ OT = OT \quad \text{(Common)} \]
\[ \therefore \Delta OVT \cong \Delta OUT \quad \text{(RHS congruence rule)} \]
\[ \Rightarrow \angle OTV = \angle OTU \quad \text{(By CPCT)} \]
Hence, the line joining the point of intersection to the centre makes equal angles with the chords.
Q4
If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that \( AB = CD \).
Solution:
Let us draw a perpendicular OM on line AD.
BC is the chord of the smaller circle and AD is the chord of the bigger circle.
Perpendicular drawn from the centre bisects the chord:
\[ BM = MC \quad \dots(1) \]
\[ AM = MD \quad \dots(2) \]
Subtracting equation (1) from (2):
\[ AM - BM = MD - MC \]
\[ \Rightarrow AB = CD \]
Q5
Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m. 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:
Draw perpendiculars OA and OB on RS and SM respectively.
\[ AR = AS = \frac{6}{2} = 3 \text{ m} \]
\[ OR = OS = OM = 5 \text{ m} \]
In \( \Delta OAR \):
\[ OA^2 + AR^2 = OR^2 \]
\[ OA^2 + 9 = 25 \Rightarrow OA^2 = 16 \Rightarrow OA = 4 \text{ m} \]
Quadrilateral ORSM is a kite (\( OR=OM \) and \( RS=SM \)). Diagonals are perpendicular. Let diagonals intersect at C.
Area of \( \Delta ORS = \frac{1}{2} \times OA \times RS = \frac{1}{2} \times 4 \times 6 = 12 \text{ m}^2 \).
Also, Area of \( \Delta ORS = \frac{1}{2} \times OS \times RC \).
\[ 12 = \frac{1}{2} \times 5 \times RC \Rightarrow RC = \frac{24}{5} = 4.8 \text{ m} \]
\[ RM = 2 \times RC = 2 \times 4.8 = 9.6 \text{ m} \]
Q6
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:
Given \( AS = SD = DA \), so \( \Delta ASD \) is equilateral. Radius \( OA = 20 \) m.
The median AB passes through the centre O. Centroid divides median in 2:1 ratio.
\[ \frac{OA}{OB} = \frac{2}{1} \Rightarrow \frac{20}{OB} = 2 \Rightarrow OB = 10 \text{ m} \]
Total height \( AB = 20 + 10 = 30 \text{ m} \).
In \( \Delta ABD \), let side be \( a \). Then \( BD = a/2 \).
\[ AD^2 = AB^2 + BD^2 \]
\[ a^2 = 30^2 + (a/2)^2 \Rightarrow a^2 - \frac{a^2}{4} = 900 \]
\[ \frac{3a^2}{4} = 900 \Rightarrow a^2 = 1200 \Rightarrow a = 20\sqrt{3} \]
The string length is \( 20\sqrt{3} \) m.
Exercise 9.3
Q1
In the given figure, A, B and C are three points on a circle with centre O such that \( \angle BOC = 30^{\circ} \) and \( \angle AOB = 60^{\circ} \). If D is a point on the circle other than the arc ABC, find \( \angle ADC \).
Solution:
\[ \angle AOC = \angle AOB + \angle BOC = 60^{\circ} + 30^{\circ} = 90^{\circ} \]
Angle at the centre is double the angle at the remaining part of the circle.
\[ \angle ADC = \frac{1}{2} \angle AOC = \frac{1}{2} (90^{\circ}) = 45^{\circ} \]
Q2
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 \( \Delta OAB \), \( AB = OA = OB \). Thus, \( \Delta OAB \) is equilateral.
\[ \angle AOB = 60^{\circ} \]
Angle on the major arc (at point C):
\[ \angle ACB = \frac{1}{2} \angle AOB = 30^{\circ} \]
Angle on the minor arc (at point D) in cyclic quadrilateral ACBD:
\[ \angle ACB + \angle ADB = 180^{\circ} \]
\[ 30^{\circ} + \angle ADB = 180^{\circ} \Rightarrow \angle ADB = 150^{\circ} \]
Q3
In the given figure, \( \angle PQR = 100^{\circ} \), where P, Q and R are points on a circle with centre O. Find \( \angle OPR \).
Solution:
Take point S on the major arc. PQRS is a cyclic quadrilateral.
\[ \angle PQR + \angle PSR = 180^{\circ} \Rightarrow 100^{\circ} + \angle PSR = 180^{\circ} \Rightarrow \angle PSR = 80^{\circ} \]
Angle at center is double the angle at the arc:
\[ \angle POR = 2 \angle PSR = 160^{\circ} \]
In \( \Delta POR \), \( OP = OR \) (radii). Thus \( \angle OPR = \angle ORP \).
\[ 2 \angle OPR + 160^{\circ} = 180^{\circ} \]
\[ 2 \angle OPR = 20^{\circ} \Rightarrow \angle OPR = 10^{\circ} \]
Q4
In the given figure, \( \angle ABC = 69^{\circ} \), \( \angle ACB = 31^{\circ} \) find \( \angle BDC \).
Solution:
In \( \Delta ABC \):
\[ \angle BAC = 180^{\circ} - (\angle ABC + \angle ACB) = 180^{\circ} - (69^{\circ} + 31^{\circ}) = 180^{\circ} - 100^{\circ} = 80^{\circ} \]
Angles in the same segment are equal:
\[ \angle BDC = \angle BAC = 80^{\circ} \]
Q5
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 \( \angle BEC = 130^{\circ} \) and \( \angle ECD = 20^{\circ} \). Find \( \angle BAC \).
Solution:
In \( \Delta CDE \), \( \angle BEC \) is the exterior angle.
\[ \angle CDE + \angle DCE = \angle BEC \]
\[ \angle CDE + 20^{\circ} = 130^{\circ} \Rightarrow \angle CDE = 110^{\circ} \]
Angles in the same segment are equal:
\[ \angle BAC = \angle CDE = 110^{\circ} \]
Q6
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If \( \angle DBC = 70^{\circ} \), \( \angle BAC \) is \( 30^{\circ} \), find \( \angle BCD \). Further, if \( AB = BC \), find \( \angle ECD \).
Solution:
Angles in the same segment are equal: \( \angle CAD = \angle DBC = 70^{\circ} \).
\[ \angle BAD = \angle BAC + \angle CAD = 30^{\circ} + 70^{\circ} = 100^{\circ} \]
Opposite angles of cyclic quadrilateral sum to \( 180^{\circ} \):
\[ \angle BCD + \angle BAD = 180^{\circ} \Rightarrow \angle BCD = 180^{\circ} - 100^{\circ} = 80^{\circ} \]
If \( AB = BC \), then \( \angle BCA = \angle BAC = 30^{\circ} \).
\[ \angle ECD = \angle BCD - \angle BCA = 80^{\circ} - 30^{\circ} = 50^{\circ} \]
Q7
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 the cyclic quadrilateral. Diagonals AC and BD are diameters.
Angle subtended by a diameter at the semicircle is \( 90^{\circ} \).
\[ \angle DAB = 90^{\circ}, \quad \angle ABC = 90^{\circ}, \quad \angle BCD = 90^{\circ}, \quad \angle CDA = 90^{\circ} \]
Since all interior angles are \( 90^{\circ} \), ABCD is a rectangle.
Q8
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Solution:
Trapezium ABCD with \( AB \parallel CD \) and \( AD = BC \). Draw perpendiculars \( AM \perp CD \) and \( BN \perp CD \).
In \( \Delta AMD \) and \( \Delta BNC \):
\[ AD = BC \quad \text{(Given)} \]
\[ \angle AMD = \angle BNC = 90^{\circ} \]
\[ AM = BN \quad \text{(Distance between parallel lines)} \]
\[ \therefore \Delta AMD \cong \Delta BNC \quad \text{(RHS)} \]
\[ \Rightarrow \angle ADC = \angle BCD \]
Also \( \angle BAD + \angle ADC = 180^{\circ} \) (Interior angles). Thus \( \angle BAD + \angle BCD = 180^{\circ} \).
Since opposite angles are supplementary, ABCD is cyclic.
Q9
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. Prove that \( \angle ACP = \angle QCD \).
Solution:
For the first circle, angles in the same segment: \( \angle ACP = \angle ABP \).
For the second circle, angles in the same segment: \( \angle QCD = \angle QBD \).
Vertically opposite angles: \( \angle ABP = \angle QBD \).
Therefore, \( \angle ACP = \angle QCD \).
Q10
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 \( \Delta ABC \). Circles on AB and AC as diameters intersect at D. Join AD.
Angle in a semicircle is \( 90^{\circ} \).
\[ \angle ADB = 90^{\circ} \]
\[ \angle ADC = 90^{\circ} \]
\[ \angle BDC = \angle ADB + \angle ADC = 180^{\circ} \]
Therefore, BDC is a straight line, which implies D lies on BC.
Q11
ABC and ADC are two right triangles with common hypotenuse AC. Prove that \( \angle CAD = \angle CBD \).
Solution:
Since \( \angle B = 90^{\circ} \) and \( \angle D = 90^{\circ} \), the sum \( \angle B + \angle D = 180^{\circ} \).
Thus, ABCD is a cyclic quadrilateral with AC as diameter.
Angles in the same segment (subtended by chord CD) are equal:
\[ \angle CAD = \angle CBD \]
Q12
Prove that a cyclic parallelogram is a rectangle.
Solution:
Let ABCD be a cyclic parallelogram.
Opposite angles of a parallelogram are equal: \( \angle A = \angle C \).
Opposite angles of a cyclic quadrilateral sum to \( 180^{\circ} \): \( \angle A + \angle C = 180^{\circ} \).
\[ \angle A + \angle A = 180^{\circ} \Rightarrow 2 \angle A = 180^{\circ} \Rightarrow \angle A = 90^{\circ} \]
Since one angle is \( 90^{\circ} \), the parallelogram is a rectangle.