CBSE Class 9 Mathematics Circles Worksheet Set 02

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Question 1. It is possible to draw a circle which passes through three collinear points (T/F)
Answer: False.
If three points lie on the exact same straight line, the perpendicular bisectors of the segments joining them are parallel and will never intersect. Because there is no common point equidistant from all three points, no circle can pass through three collinear points.
In simple words: A circle bends, so it cannot pass through three points sitting in a straight row.

Exam Tip: Remember the theorem: exactly one unique circle passes through three non-collinear points, whereas zero circles can pass through three collinear points.

Question 2. The perpendicular bisector of two chords intersect at centre of circle (T/F)
Answer: True.
The perpendicular bisector of every chord of a circle always passes through its centre. Therefore, the perpendicular bisectors of any two non-parallel chords must cross each other at the circle's centre.
In simple words: Cutting any two chords down the middle at right angles always points straight to the center.

Exam Tip: This property is the fundamental method used to locate the unknown centre of any given circle or arc.

Question 3. If two arcs of a circle are congruent. Then corresponding chords are unequal (T/F)
Answer: False.
Congruent arcs of a circle have identical degree measures and equal lengths. Consequently, the straight-line segments connecting their endpoints must be equal in length.
In simple words: Arcs of the same size always produce chords of the exact same length.

Exam Tip: Remember the bi-conditional rule: congruent arcs have equal chords, and equal chords subtend congruent arcs.

Question 4. The line joining the mid-point of a chord to centre of circle is perpendicular to chord. (T/F)
Answer: True.
By standard circle theorem, joining the centre of a circle to the midpoint of a non-diameter chord creates two congruent triangles with the radii, ensuring the joining line meets the chord at a \( 90^\circ \) angle.
In simple words: A line drawn from the center to the exact middle of a chord always hits it at a right angle.

Exam Tip: Quote this directly as Theorem 10.4: "The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord."

Question 5. If O is the center of circle of radius 5 cm OP perpendicular to AB and OQ perpendicular to CD, AB||CD, AB = 6cm and CD = 8 cm. Determine PQ.
Answer: The perpendicular drawn from the centre to a chord bisects that chord.
For chord \( AB = 6\text{ cm} \), the half-length is \( 3\text{ cm} \). In right-angled triangle \( \Delta OPA \):
\( OP = \sqrt{OA^2 - AP^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm} \)
For chord \( CD = 8\text{ cm} \), the half-length is \( 4\text{ cm} \). In right-angled triangle \( \Delta OQC \):
\( OQ = \sqrt{OC^2 - CQ^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\text{ cm} \)
Depending on the relative positions of the parallel chords with respect to centre \( O \):
- Case 1: If chords lie on opposite sides of the centre, \( PQ = OP + OQ = 4 + 3 = 7\text{ cm} \).
- Case 2: If chords lie on the same side of the centre, \( PQ = OP - OQ = 4 - 3 = 1\text{ cm} \).
In simple words: Using Pythagoras theorem gives distances of 4 cm and 3 cm from the center. If they sit on opposite sides, the distance is 7 cm; if on the same side, it is 1 cm.

Exam Tip: Always evaluate both possibilities (same side vs. opposite sides of the centre) whenever the orientation of the parallel chords is not explicitly specified.

Question 6. In the given figure what is the measurement of angle ABC
(a) 50°
(b) 40°
(c) 90°
(d) 75°
Answer: (b) 40°
According to the degree measure theorem, the angle subtended by an arc at the centre of a circle is twice the angle subtended by it at any point on the remaining circumference:
\( \angle ABC = \frac{1}{2} \angle AOC = \frac{1}{2} \times 80^\circ = 40^\circ \)
In simple words: The angle at the top edge is half of the central angle, so half of 80° gives 40°.

Exam Tip: Verify which arc subtends the angles before halving: the central angle and circumference angle must be subtended by the same arc.

Question 7. Find the angle ADC
(a) 20 degree
(b) 40 degree
(c) 60 degree
(d) 80 degree
Answer: (b) 40 degree
Both angles \( \angle ABC \) and \( \angle ADC \) are subtended by the same chord \( AC \) and lie in the same circular segment. By circle theorem, angles in the same segment of a circle are equal:
\( \angle ADC = \angle ABC = 40^\circ \)
In simple words: Since both angles stand on the same base chord AC, they are identical, making angle ADC equal to 40°.

Exam Tip: State the justification clearly: "Angles in the same segment of a circle are equal."

Question 8. The length of the chord of a circle is 30 cm and its distance from the centre is 8 cm. Find the radius of the circle.
(a) 10 cm
(b) 20 cm
(c) 34 cm
(d) 17 cm.
Answer: (d) 17 cm.
The perpendicular dropped from the centre bisects the chord, giving a half-length of \( \frac{30}{2} = 15\text{ cm} \).
Applying the Pythagorean theorem in the right-angled triangle formed by the radius, perpendicular distance, and half-chord:
\( r = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17\text{ cm} \)
In simple words: Half the chord is 15 cm. Combining 15 cm and 8 cm with Pythagoras gives a radius of 17 cm.

Exam Tip: Recognizing the common Pythagorean triple (8, 15, 17) allows you to solve and verify this calculation in seconds.

Question 9. In the given figure, chords AB and CD subtend equal angles each = 70° at the centre O of the circle. OL ⊥ AB and AL = 2.5 cm find CD.
(a) 2 cm
(b) 2.5 cm
(c) 7.5 cm
(d) 5 cm
Answer: (d) 5 cm
Because chords subtending equal angles at the centre of a circle are equal in length:
\( AB = CD \)
Since \( OL \perp AB \), the line from the centre bisects the chord:
\( AB = 2 \times AL = 2 \times 2.5 = 5\text{ cm} \)
Consequently, \( CD = 5\text{ cm} \).
In simple words: The full length of AB is twice 2.5 cm, which is 5 cm. Since CD subtends the same angle at the center, CD is also 5 cm.

Exam Tip: First double the bisected segment to get the full chord, then apply the theorem that chords subtending equal central angles are equal.

Question 10. The adjoining figure, AB is the diameter of the circle with centre O. A chord CD is bisected by the diameter at P. If OA = OB = 15 cm and OP = 9 cm. Find the length of chord AD.
(a) 12√5 cm
(b) 12 cm
(c) 13cm
(d) 13√5cm
Answer: (a) 12√5 cm
Because diameter \( AB \) bisects chord \( CD \) at \( P \), \( AB \perp CD \), making \( \angle OPD = 90^\circ \).
In right triangle \( \Delta OPD \), with radius \( OD = 15\text{ cm} \) and \( OP = 9\text{ cm} \):
\( PD = \sqrt{OD^2 - OP^2} = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12\text{ cm} \)
Now compute segment \( AP \):
\( AP = OA + OP = 15 + 9 = 24\text{ cm} \)
In right triangle \( \Delta APD \):
\( AD = \sqrt{AP^2 + PD^2} = \sqrt{24^2 + 12^2} = \sqrt{576 + 144} = \sqrt{720} = 12\sqrt{5}\text{ cm} \)
In simple words: Pythagoras on triangle OPD gives half-chord PD = 12 cm. Then on triangle APD with base 24 cm and height 12 cm, AD works out to 12√5 cm.

Exam Tip: Make sure to add \( OA + OP = 15 + 9 = 24\text{ cm} \) to obtain the full length of leg \( AP \) before applying Pythagoras theorem in \( \Delta APD \).

Question 11. In a circle radius is 13cm, a chord is drawn at a distance of 12cm from the centre. Find the length of the chord.
(a) 5 cm
(b) 10 cm
(c) 15 cm
(d) 30 cm
Answer: (b) 10 cm
Let half the length of the chord be \( x \). In the right-angled triangle formed with the radius:
\( x = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5\text{ cm} \)
The total length of the chord is double this value:
\( \text{Length of chord} = 2 \times 5 = 10\text{ cm} \)
In simple words: Pythagoras theorem gives 5 cm for half the chord, so the whole chord is 10 cm.

Exam Tip: Do not stop at the half-chord length of 5 cm; always multiply by 2 to state the full chord length.

Question 12. Two parallel chords of a circle whose radius is 13 cm are respectively 10 cm and 24 cm. If they live on the opposite side of the centre find the distance between them.
(a) 10 cm
(b) 7 cm
(c) 17 cm
(d) 13 cm
Answer: (c) 17 cm
Half-lengths of the two chords are \( \frac{10}{2} = 5\text{ cm} \) and \( \frac{24}{2} = 12\text{ cm} \).
Distance of the 10 cm chord from the centre:
\( d_1 = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm} \)
Distance of the 24 cm chord from the centre:
\( d_2 = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5\text{ cm} \)
Because the chords lie on opposite sides of the centre, the total distance between them is the sum of their individual distances:
\( \text{Distance} = d_1 + d_2 = 12 + 5 = 17\text{ cm} \)
In simple words: The distances from the center are 12 cm and 5 cm. Because they sit on opposite sides, add them together to get 17 cm.

Exam Tip: Add the distances when chords are on opposite sides of the centre, and subtract them when chords are on the same side.

Question 13. In the adjoining figure, N is the mid point of the chord LM of a circle where centre is O. If ∠OLN = 40°, find the measure of ∠LON
(a) 50°
(b) 120°
(c) 90°
(d) 75°
Answer: (a) 50°
Because \( N \) is the midpoint of chord \( LM \), the segment from the centre is perpendicular to the chord:
\( ON \perp LM \implies \angle ONL = 90^\circ \)
In right-angled triangle \( \Delta OLN \):
\( \angle LON = 90^\circ - \angle OLN = 90^\circ - 40^\circ = 50^\circ \)
In simple words: The triangle contains a 90° right angle at N and 40° at L, so the third angle at the center is 50°.

Exam Tip: Recall that the acute angles of any right-angled triangle always sum to \( 90^\circ \).

Question 14. Two parallel chords of a circle whose radius is √117 cm, are 12 cm and 18 cm. Find the distance between them. Both chords are on the same side of the centre.
(a) 7 cm
(b) 5 cm
(c) 3 cm
(d) 15 cm
Answer: (c) 3 cm
Half-lengths of the chords are \( \frac{18}{2} = 9\text{ cm} \) and \( \frac{12}{2} = 6\text{ cm} \).
Perpendicular distance to the 18 cm chord:
\( d_1 = \sqrt{(\sqrt{117})^2 - 9^2} = \sqrt{117 - 81} = \sqrt{36} = 6\text{ cm} \)
Perpendicular distance to the 12 cm chord:
\( d_2 = \sqrt{(\sqrt{117})^2 - 6^2} = \sqrt{117 - 36} = \sqrt{81} = 9\text{ cm} \)
Because both chords reside on the same side of the centre, subtract the distances:
\( \text{Distance} = d_2 - d_1 = 9 - 6 = 3\text{ cm} \)
In simple words: The two distances from the center are 9 cm and 6 cm. Since they lie on the same side, their difference is 3 cm.

Exam Tip: Notice that the longer chord is always closer to the centre of the circle.

Question 15. In figure, a square is inscribed in a circle of radius 8√2 cm. Find the length of the square
(a) 12 cm
(b) 8 cm
(c) 16 cm
(d) 10 cm
Answer: (c) 16 cm
The diagonal of a square inscribed inside a circle is equal to the circle's diameter:
\( \text{Diagonal } d = 2 \times 8\sqrt{2} = 16\sqrt{2}\text{ cm} \)
The relationship between side length \( s \) and diagonal \( d \) of a square is \( d = s\sqrt{2} \):
\( s = \frac{d}{\sqrt{2}} = \frac{16\sqrt{2}}{\sqrt{2}} = 16\text{ cm} \)
In simple words: The diameter is 16√2 cm. Dividing the diagonal by √2 gives a side length of 16 cm.

Exam Tip: For any square inscribed in a circle of radius \( R \), the side length is always given by \( s = R\sqrt{2} \).

Question 16. In the given figure, O is the centre of the circle, AB and CD are two chords such that OL is perpendicular to AB and OM is perpendicular to CD. ∠AOB = 50°, AL = DM = 2 cm. Find the measure of ∠COD.
(a) 50°
(b) 70°
(c) 90°
(d) 130°
Answer: (a) 50°
Because the perpendicular from the centre bisects each chord:
\( AB = 2 \times AL = 2 \times 2 = 4\text{ cm} \)
\( CD = 2 \times DM = 2 \times 2 = 4\text{ cm} \)
Since chords \( AB \) and \( CD \) are equal in length, they subtend equal angles at the centre:
\( \angle COD = \angle AOB = 50^\circ \)
In simple words: Both chords are 4 cm long, and equal chords always create equal angles at the center (50°).

Exam Tip: State Theorem 10.1: "Equal chords of a circle subtend equal angles at the centre."

Question 17. ΔABC is inscribed in a circle with centre O. IF ∠ACB = 40°, find ∠B.
(a) 65°
(b) 90°
(c) 40°
(d) 50°
Answer: (d) 50°
Because segment \( BC \) passes through centre \( O \), it is a diameter. The angle inscribed in a semicircle is a right angle:
\( \angle BAC = 90^\circ \)
In \( \Delta ABC \), the angles sum to \( 180^\circ \):
\( \angle B = 180^\circ - (90^\circ + 40^\circ) = 50^\circ \)
In simple words: The diameter forms a 90° angle at corner A. Subtracting 90° and 40° from 180° leaves 50° for angle B.

Exam Tip: Whenever a chord passes through the centre, identify it as a diameter and immediately mark the opposite inscribed angle as \( 90^\circ \).

Question 18. ΔABC is inscribed in a circle and AD ⊥ BC meets BC at D and AD produced meets the circle at L. If ∠ABD is 35° then find the measure of ∠BCL.
(a) 55°
(b) 60°
(c) 90°
(d) 35°
Answer: (a) 55°
In right-angled triangle \( \Delta ABD \), \( \angle ADB = 90^\circ \) and \( \angle ABD = 35^\circ \):
\( \angle BAD = 90^\circ - 35^\circ = 55^\circ \)
Since line segment \( ADL \) is straight, \( \angle BAL = \angle BAD = 55^\circ \).
Both \( \angle BAL \) and \( \angle BCL \) are subtended by the same arc \( BL \) in the same circular segment:
\( \angle BCL = \angle BAL = 55^\circ \)
In simple words: The right triangle gives 55° at corner A. Because angle A and angle BCL stand on the same arc BL, angle BCL is also 55°.

Exam Tip: Trace the common arc \( BL \) connecting vertices \( A \) and \( C \) to easily spot angles in the same segment.

Question 19. In the adjoining figure, O is the centre of the circle, find the measure of angle X.
(a) 85°
(b) 110°
(c) 50°
(d) 35°
Answer: (c) 50°
Because \( AC \) and \( BD \) are diameters intersecting at centre \( O \):
In \( \Delta OAD \), \( OA = OD \) (radii), so \( \angle OAD = \angle ODA = 50^\circ \).
The angle at the centre is \( \angle AOD = 180^\circ - (50^\circ + 50^\circ) = 80^\circ \).
Vertically opposite angle \( \angle BOC = \angle AOD = 80^\circ \).
In \( \Delta OBC \), \( OB = OC \) (radii), so \( \angle OBC = \angle OCB = X^\circ \):
\( 2X + 80^\circ = 180^\circ \implies 2X = 100^\circ \implies X = 50^\circ \)
In simple words: The radius sides make both triangles isosceles. The vertically opposite angles match, so corner X equals 50°.

Exam Tip: Radii drawn from the centre to any points on the circle always form isosceles triangles with equal base angles.

Question 20. In the figure, ∠Q = 80° find ∠S
(a) 100°
(b) 80°
(c) 90°
(d) 110°
Answer: (a) 100°
Quadrilateral \( PQRS \) is inscribed in the circle, making it a cyclic quadrilateral.
The opposite angles of any cyclic quadrilateral are supplementary:
\( \angle Q + \angle S = 180^\circ \)
\( 80^\circ + \angle S = 180^\circ \)
\( \implies \angle S = 180^\circ - 80^\circ = 100^\circ \)
In simple words: Opposite corners of a cyclic quadrilateral add up to 180°, so 180° minus 80° gives 100°.

Exam Tip: State the cyclic quadrilateral theorem directly: "The opposite angles of a cyclic quadrilateral are supplementary."

Question 21. From the given figure, find the value of x
(a) 108°
(b) 72°
(c) 60°
(d) 36°
Answer: (d) 36°
Because all four vertices of quadrilateral \( ABCD \) lie on the circle, it is a cyclic quadrilateral.
Opposite angles are supplementary:
\( \angle A + \angle C = 180^\circ \)
\( 2x + 3x = 180^\circ \)
\( \implies 5x = 180^\circ \)
\( \implies x = 36^\circ \)
In simple words: The opposite angles add to 5x. Dividing 180° by 5 gives x = 36°.

Exam Tip: Combine the variable coefficients first (\( 2x + 3x = 5x \)) before dividing by 5.

Question 22. In the given figure, O is the centre of the circle. If ∠AOC = 90°, find ∠ABC
(a) 45°
(b) 90°
(c) 135°
(d) 75°
Answer: (c) 135°
Take point \( D \) on the major arc \( AC \). The angle subtended by minor arc \( AC \) at point \( D \) is half the central angle:
\( \angle ADC = \frac{1}{2} \angle AOC = \frac{1}{2} \times 90^\circ = 45^\circ \)
Quadrilateral \( ABCD \) is a cyclic quadrilateral, so its opposite angles are supplementary:
\( \angle ABC + \angle ADC = 180^\circ \)
\( \angle ABC = 180^\circ - 45^\circ = 135^\circ \)
In simple words: The angle on the opposite side of the circle is half of 90°, which is 45°. Because opposite angles in a cyclic shape sum to 180°, angle ABC is 180° - 45° = 135°.

Exam Tip: For reflex arcs, you can also use the reflex angle directly: \( \angle ABC = \frac{1}{2} (\text{Reflex } \angle AOC) = \frac{1}{2}(360^\circ - 90^\circ) = 135^\circ \).

Question 23. Given cyclic trapezium ABCD in which AD || BC and ∠B = 60°, Determine ∠A.
(a) 90°
(b) 120°
(c) 60°
(d) 30°
Answer: (b) 120°
Because \( AD \parallel BC \) and line \( AB \) serves as a transversal intersecting these parallel sides, consecutive interior angles on the same side of the transversal are supplementary:
\( \angle A + \angle B = 180^\circ \)
\( \angle A + 60^\circ = 180^\circ \)
\( \implies \angle A = 180^\circ - 60^\circ = 120^\circ \)
In simple words: Parallel lines make co-interior angles add up to 180°, so 180° minus 60° gives 120° for angle A.

Exam Tip: Use the parallel lines property (\( \angle A + \angle B = 180^\circ \)) directly rather than finding the opposite cyclic angle first.

Question 24. In the adjoining figure, in circle with centre O. ∠BDC = 40°, find ∠ACB.
(a) 50°
(b) 45°
(c) 40°
(d) 90°
Answer: (d) 90°
In the circle with centre \( O \), line segment \( AB \) passes through the centre and is therefore a diameter.
By circle theorem, the angle subtended by a diameter in a semicircle is a right angle:
\( \angle ACB = 90^\circ \)
(Note: If the question intended to find \( \angle ABC \), then since \( \angle BAC = \angle BDC = 40^\circ \), \( \angle ABC = 90^\circ - 40^\circ = 50^\circ \)).
In simple words: The line AB goes through the center, making it a diameter. Any angle formed on the rim by a diameter is always 90°.

Exam Tip: Remember: "Angle in a semicircle is a right angle (\( 90^\circ \))."

Question 25. ABC is inscribed in a circle. The bisector of ∠A meets the circle in L and the bisector of ∠C meets AL at K. If ∠BAC = 80° and ∠ACB = 60°, find ∠LCK
(a) 40°
(b) 30°
(c) 90°
(d) 70°
Answer: (d) 70°
In \( \Delta ABC \), \( \angle B = 180^\circ - (80^\circ + 60^\circ) = 40^\circ \).
Because \( AL \) bisects \( \angle BAC \):
\( \angle BAL = \angle CAL = \frac{80^\circ}{2} = 40^\circ \)
Angles in the same segment subtended by chord \( AB \) are equal:
\( \angle CLK = \angle CLA = \angle ABC = 40^\circ \)
Because \( CK \) bisects \( \angle ACB \):
\( \angle ACK = \frac{60^\circ}{2} = 30^\circ \)
In \( \Delta ACK \), exterior angle \( \angle CKL \) equals the sum of interior opposite angles:
\( \angle CKL = \angle KAC + \angle ACK = 40^\circ + 30^\circ = 70^\circ \)
In \( \Delta CKL \), the angles sum to \( 180^\circ \):
\( \angle LCK = 180^\circ - (\angle CLK + \angle CKL) = 180^\circ - (40^\circ + 70^\circ) = 70^\circ \)
In simple words: The exterior angle to the triangle is 40° + 30° = 70°. In the lower triangle CKL, the angles sum to 180°, leaving 70° for angle LCK.

Exam Tip: Use the exterior angle theorem on \( \Delta ACK \) to find \( \angle CKL = 70^\circ \) quickly.

Question 26. In the adjoining figure, O is the centre of the circle, ∠AOB = 90°, ∠BOC = 120°
(a) 150°
(b) 75°
(c) 50°
(d) 120°
Answer: (b) 75°
(Note: The question asks to find the measure of \( \angle ABC \)).
The sum of all central angles surrounding centre \( O \) equals \( 360^\circ \):
\( \angle AOC = 360^\circ - (\angle AOB + \angle BOC) = 360^\circ - (90^\circ + 120^\circ) = 360^\circ - 210^\circ = 150^\circ \)
By the degree measure theorem, the angle subtended by arc \( AC \) at circumference vertex \( B \) is half the central angle:
\( \angle ABC = \frac{1}{2} \angle AOC = \frac{1}{2} \times 150^\circ = 75^\circ \)
In simple words: The remaining angle at the center is 360° minus 210°, giving 150°. Angle ABC on the circumference is half of that, which equals 75°.

Exam Tip: First compute the central angle \( \angle AOC \) subtended by arc \( AC \), then halve it to find \( \angle ABC \).

Question 27. In the figure, diameter AB = 30 cm, chord AC = 24 cm. Find how far is the chord AC from the centre of the circle ?
(a) 9 cm
(b) 8√5 cm
(c) 6 cm
(d) 12 cm
Answer: (a) 9 cm
Radius of the circle \( r = \frac{\text{Diameter}}{2} = \frac{30}{2} = 15\text{ cm} \).
The perpendicular from the centre bisects chord \( AC \), giving half-length \( \frac{24}{2} = 12\text{ cm} \).
In the right-angled triangle formed by the radius, distance \( d \), and half-chord:
\( d = \sqrt{15^2 - 12^2} = \sqrt{225 - 144} = \sqrt{81} = 9\text{ cm} \)
In simple words: The radius is 15 cm and half the chord is 12 cm. Pythagoras theorem gives a distance of 9 cm.

Exam Tip: Be sure to divide the diameter by 2 first to obtain the radius of 15 cm before using Pythagoras theorem.

Question 28. In the adjoining figure, CD is the diameter which meets the chord AB in E such that AE = BE = 4 cm. If CE is 3 cm. Find the radius of the circle.
(a) 4 1/6 cm
(b) 2 1/7 cm
(c) 3 cm
(d) 4 cm
Answer: (a) 4 1/6 cm
Let the radius of the circle be \( r \).
Because diameter \( CD \) bisects chord \( AB \) at \( E \), \( CD \perp AB \), and \( OE = OC - CE = r - 3 \).
In right-angled triangle \( \Delta OEB \), with hypotenuse \( OB = r \) and \( BE = 4\text{ cm} \):
\( OB^2 = OE^2 + BE^2 \)
\( r^2 = (r - 3)^2 + 4^2 \)
\( r^2 = r^2 - 6r + 9 + 16 \)
\( \implies 6r = 25 \)
\( \implies r = \frac{25}{6} = 4\frac{1}{6}\text{ cm} \)
In simple words: Setting up the Pythagoras equation r² = (r - 3)² + 4² cancels out r² and leaves 6r = 25, which gives r = 4 1/6 cm.

Exam Tip: Express \( OE \) as \( r - CE \) to form a clean linear equation in \( r \) once \( r^2 \) cancels out.

Question 29. In the given figure AOB and COD are the two diameters of a circle with centre O. If ∠BDO = 50°, find ∠ACO
(a) 50°
(b) 130°
(c) 90°
(d) 75°
Answer: (a) 50°
In \( \Delta OBD \), \( OB = OD \) (radii of the circle):
\( \angle OBD = \angle BDO = 50^\circ \)
Because \( AOB \) and \( COD \) are diameters intersecting at centre \( O \), line segments \( AC \) and \( BD \) connect endpoints of intersecting diameters, making \( AC \parallel BD \).
Transversal diameter \( COD \) cuts parallel lines \( AC \) and \( BD \), so alternate interior angles are equal:
\( \angle ACO = \angle BDO = 50^\circ \)
In simple words: The lines AC and BD are parallel. Alternate interior angles on transversal CD make angle ACO match 50°.

Exam Tip: Use alternate interior angles directly since chords joining opposite endpoints of two diameters are always parallel.

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Question 30. O is the centre of a circle and the measure of arc ABC is 100 degree. Determine angle ADC and Angle ABC
Answer: The central angle subtended by minor arc \( ABC \) is \( \angle AOC = 100^\circ \).
By the degree measure theorem, the angle subtended by this arc at any point \( D \) on the remaining circumference is half the central angle:
\( \angle ADC = \frac{1}{2} \angle AOC = \frac{1}{2} \times 100^\circ = 50^\circ \)
Quadrilateral \( ABCD \) is inscribed in the circle, which makes it a cyclic quadrilateral.
Opposite angles of a cyclic quadrilateral are supplementary:
\( \angle ABC + \angle ADC = 180^\circ \)
\( \angle ABC = 180^\circ - 50^\circ = 130^\circ \)
Therefore, \( \angle ADC = 50^\circ \) and \( \angle ABC = 130^\circ \).
In simple words: Half of the 100° central angle gives angle ADC = 50°. Then subtracting 50° from 180° gives angle ABC = 130°.

Exam Tip: Clearly show the two steps: finding \( \angle ADC \) using the central angle theorem first, then using the cyclic quadrilateral property for \( \angle ABC \).

Question 31. A line l intersecting two concurrent circles having same center is O, at the point A,B,C,D. Show that AB=CD.
Answer: Let a line \( l \) intersect the outer concentric circle at points \( A \) and \( D \), and the inner concentric circle at points \( B \) and \( C \).
From common centre \( O \), drop a perpendicular \( OM \perp l \).
According to circle theorem, the perpendicular from the centre of a circle to a chord bisects the chord:
- For chord \( AD \) of the larger circle:
\( AM = MD \) ...(1)
- For chord \( BC \) of the smaller circle:
\( BM = MC \) ...(2)
Subtracting equation (2) from equation (1):
\( AM - BM = MD - MC \)
\( \implies AB = CD \)
Hence proved.
In simple words: Drop a perpendicular from the center onto the line. It cuts both the long chord and the short chord in half. Subtracting the short half from the long half leaves AB = CD.

Exam Tip: State the construction "Draw \( OM \perp l \)" explicitly to earn the designated method mark.

Question 32. Prove that the line joining to the centre of circle to the mid-point of a chord, is perpendicular to the chord.
Answer: Let \( AB \) be a chord of a circle with centre \( O \), and let \( M \) be the midpoint of chord \( AB \), so \( AM = BM \).
Join line segments \( OA \), \( OB \), and \( OM \).
In \( \Delta OMA \) and \( \Delta OMB \):
- \( OA = OB \) (radii of the same circle)
- \( AM = BM \) (since \( M \) is the midpoint of \( AB \))
- \( OM = OM \) (common side)
By the SSS congruence criterion:
\( \Delta OMA \cong \Delta OMB \)
By CPCT:
\( \angle OMA = \angle OMB \)
Because \( AMB \) is a straight line segment, these adjacent angles form a linear pair:
\( \angle OMA + \angle OMB = 180^\circ \)
\( \angle OMA + \angle OMA = 180^\circ \)
\( \implies 2\angle OMA = 180^\circ \)
\( \implies \angle OMA = 90^\circ \)
Therefore, \( OM \perp AB \).
Hence proved.
In simple words: Joining the center to the chord ends forms two identical triangles by SSS. Because their angles at the midpoint are equal and lie on a flat line, each must be 90°.

Exam Tip: Remember to use the linear pair property (\( \angle OMA + \angle OMB = 180^\circ \)) after proving congruence to establish the right angle.

Question 33. Prove that equal chords of a circle subtend equal angles at the centre.
Answer: Let \( AB \) and \( CD \) be two equal chords of a circle with centre \( O \), so \( AB = CD \).
Join segments \( OA \), \( OB \), \( OC \), and \( OD \).
In \( \Delta AOB \) and \( \Delta COD \):
- \( OA = OC \) (radii of the same circle)
- \( OB = OD \) (radii of the same circle)
- \( AB = CD \) (given)
By the SSS congruence criterion:
\( \Delta AOB \cong \Delta COD \)
By CPCT:
\( \angle AOB = \angle COD \)
Hence, equal chords subtend equal angles at the centre.
Hence proved.
In simple words: The two triangles formed by the chords and radii have all three sides equal (SSS). Therefore, their angles at the center are equal.

Exam Tip: This is Theorem 10.1 in the NCERT syllabus; writing the three side pairs followed by SSS congruence criterion guarantees full marks.

Question 34. Bisector AD of angle BAC of Triangle ABC passes through the center O of the circumcircle of triangle ABC prove that AB = AC.
Answer: Because ray \( AD \) bisects \( \angle BAC \):
\( \angle BAD = \angle CAD \)
These two equal inscribed angles subtend chords \( BD \) and \( CD \) respectively on the circumcircle:
\( \text{Chord } BD = \text{Chord } CD \)
Given that the bisector \( AD \) passes through the circumcentre \( O \), segment \( AD \) forms a diameter of the circumcircle.
By the angle-in-a-semicircle theorem, the angle subtended by a diameter is a right angle:
\( \angle ABD = 90^\circ \quad \text{and} \quad \angle ACD = 90^\circ \)
In right-angled triangles \( \Delta ABD \) and \( \Delta ACD \):
- \( \angle ABD = \angle ACD = 90^\circ \)
- Hypotenuse \( AD = AD \) (common side)
- Side \( BD = CD \) (equal chords subtended by equal angles)
By the RHS congruence criterion:
\( \Delta ABD \cong \Delta ACD \)
By CPCT:
\( AB = AC \)
Hence proved.
In simple words: Since AD passes through the center, it acts as a diameter making angles B and C equal to 90°. RHS congruence then proves that AB and AC are equal.

Exam Tip: Explicitly note that \( AD \) passing through circumcentre \( O \) makes it a diameter, justifying the \( 90^\circ \) right angles at vertices \( B \) and \( C \).

Question 35. A,B,C,D are the four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130 degree and ∠ECD = 20 Degree. Find ∠BAC
Answer: Because \( BD \) is a straight line segment, angles \( \angle BEC \) and \( \angle DEC \) form a linear pair:
\( \angle DEC = 180^\circ - \angle BEC = 180^\circ - 130^\circ = 50^\circ \)
In \( \Delta CDE \), the angles sum to \( 180^\circ \):
\( \angle CDE = 180^\circ - (\angle DEC + \angle ECD) = 180^\circ - (50^\circ + 20^\circ) = 110^\circ \)
Thus, \( \angle BDC = \angle CDE = 110^\circ \).
Angles \( \angle BAC \) and \( \angle BDC \) are subtended by the same chord \( BC \) in the same circular segment:
\( \angle BAC = \angle BDC = 110^\circ \)
In simple words: The straight line leaves 50° inside the small triangle, so its third angle is 110°. Standing on the same arc, angle BAC also equals 110°.

Exam Tip: Use the linear pair property on line \( BD \) first, then apply the theorem that angles in the same segment of a circle are equal.

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Question 36. AB and CD are the two chord of the circle such that AB = 6 cm , CD = 12 cm and AB||CD, if the distance between AB and CD is 3 cm, find the radius of the circle.
Answer: Let the circle have radius \( r \) and centre \( O \).
Draw a perpendicular from centre \( O \) intersecting chord \( CD \) at \( M \) and chord \( AB \) at \( L \).
The perpendicular bisects each chord:
\( MD = \frac{12}{2} = 6\text{ cm} \quad \text{and} \quad AL = \frac{6}{2} = 3\text{ cm} \)
Because both chords cannot lie on opposite sides of the centre (since distance from centre to \( AB \) would exceed the 3 cm total separation), both chords must lie on the same side of the centre.
Let the distance of chord \( CD \) from the centre be \( OM = x\text{ cm} \).
Then the distance of chord \( AB \) from the centre is \( OL = x + 3\text{ cm} \).
Applying Pythagoras theorem in right triangles \( \Delta OMD \) and \( \Delta OLA \):
\( r^2 = OM^2 + MD^2 = x^2 + 6^2 = x^2 + 36 \) ...(1)
\( r^2 = OL^2 + AL^2 = (x + 3)^2 + 3^2 = x^2 + 6x + 9 + 9 = x^2 + 6x + 18 \) ...(2)
Equating expressions (1) and (2):
\( x^2 + 36 = x^2 + 6x + 18 \)
\( \implies 6x = 36 - 18 \)
\( \implies 6x = 18 \)
\( \implies x = 3\text{ cm} \)
Substitute \( x = 3 \) back into equation (1) to determine the radius:
\( r^2 = 3^2 + 36 = 9 + 36 = 45 \)
\( \implies r = \sqrt{45} = 3\sqrt{5}\text{ cm} \approx 6.71\text{ cm} \)
In simple words: Write Pythagoras equations for both chords using their distances x and x + 3. Solving them gives x = 3 cm, which leads to a radius of 3√5 cm.

Exam Tip: Clearly justify why the chords must lie on the same side: if they were on opposite sides, the distance would have to be greater than \( \sqrt{r^2 - 3^2} > 3\text{ cm} \).

Question 37. 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 to each other. Find the total length of the string.
Answer: Let the positions of Ankur, Syed, and David be represented by points \( A \), \( S \), and \( D \) respectively on the circle of radius \( R = 20\text{ m} \) centered at \( O \).
Because they sit at equal distances from one another, \( \Delta ASD \) is an equilateral triangle.
In an equilateral triangle, the circumcentre \( O \) coincides with the centroid, which divides each median in a \( 2:1 \) ratio.
Let \( AM \perp SD \) be a median meeting \( SD \) at \( M \):
\( OA = \frac{2}{3} AM \)
\( 20 = \frac{2}{3} AM \implies AM = 30\text{ m} \)
Let the side length of the equilateral triangle be \( s \). Since \( M \) is the midpoint of \( SD \), \( SM = \frac{s}{2} \).
Applying Pythagoras theorem in right-angled triangle \( \Delta AMS \):
\( AS^2 = AM^2 + SM^2 \)
\( s^2 = 30^2 + \left(\frac{s}{2}\right)^2 \)
\( s^2 - \frac{s^2}{4} = 900 \)
\( \implies \frac{3s^2}{4} = 900 \)
\( \implies s^2 = \frac{900 \times 4}{3} = 1200 \)
\( \implies s = \sqrt{1200} = 20\sqrt{3}\text{ m} \)
The length of each string is \( 20\sqrt{3}\text{ m} \).
The total length of all three strings combined is:
\( \text{Total length} = 3 \times 20\sqrt{3} = 60\sqrt{3}\text{ m} \approx 103.92\text{ m} \)
In simple words: The center divides the height in a 2 to 1 ratio, giving a full height of 30 m. Pythagoras theorem gives 20√3 m for each side, so all three strings total 60√3 m.

Exam Tip: State the centroid property clearly: "In an equilateral triangle, the circumcentre coincides with the centroid and divides the median in the ratio 2:1."

Question 38. Given an arc of circle how you will find its centre and complete the circle.
Answer: To find the centre and complete the circle from a given arc:
1. Mark three distinct points \( P \), \( Q \), and \( R \) along the given arc.
2. Draw straight line segments to form chords \( PQ \) and \( QR \).
3. Using a compass and straightedge, construct the perpendicular bisector of chord \( PQ \).
4. Similarly, construct the perpendicular bisector of chord \( QR \).
5. Let these two perpendicular bisectors intersect at point \( O \). Since the perpendicular bisector of any chord passes through the centre, point \( O \) is the required centre of the circle.
6. With \( O \) as the centre and radius equal to \( OP \) (or \( OQ \) or \( OR \)), draw the complete circle with a compass.
In simple words: Pick three points on the curve and connect them into two chords. Draw the perpendicular bisectors of both chords; where they cross is the center.

Exam Tip: List all steps of construction with clear numbering and specify that the intersection of the two perpendicular bisectors defines the centre.

Question 39. Two equal chord AB and CD of circle with center O, when produced meet at a point E, prove that BE=DE and AE=CE
Answer: Let \( AB \) and \( CD \) be two equal chords of a circle with centre \( O \), intersecting externally at point \( E \) when produced.
Draw perpendiculars \( OL \perp AB \) and \( OM \perp CD \). Join segment \( OE \).
Because equal chords are equidistant from the centre:
\( OL = OM \)
In right-angled triangles \( \Delta OLE \) and \( \Delta OME \):
- \( \angle OLE = \angle OME = 90^\circ \) (by construction)
- Hypotenuse \( OE = OE \) (common side)
- Side \( OL = OM \) (equal chords are equidistant from centre)
By the RHS congruence criterion:
\( \Delta OLE \cong \Delta OME \)
By CPCT:
\( LE = ME \) ...(1)
Because the perpendicular from the centre bisects the chord:
\( BL = \frac{1}{2} AB \quad \text{and} \quad DM = \frac{1}{2} CD \)
Since \( AB = CD \), their half-lengths are equal:
\( BL = DM \) ...(2)
Subtracting equation (2) from equation (1):
\( LE - BL = ME - DM \)
\( \implies BE = DE \)
Now add the full equal chords \( AB = CD \) to both sides:
\( BE + AB = DE + CD \)
\( \implies AE = CE \)
Hence proved.
In simple words: Right triangles OLE and OME are congruent by RHS, so LE = ME. Subtracting the equal half-chords leaves BE = DE, and adding the full chords gives AE = CE.

Exam Tip: Deduce \( BE = DE \) by subtraction first, then prove \( AE = CE \) by adding the equal chords \( AB = CD \).

Question 40. In a circle with centre O, Chords AB and CD intersect inside the circumference at E. Prove that ∠AOC+ ∠BOD = 2 ∠AEC
Answer: Join line segment \( BC \).
By the degree measure theorem, the central angle subtended by an arc is double the angle subtended by it at the circumference:
- For arc \( AC \):
\( \angle AOC = 2\angle ABC \) ...(1)
- For arc \( BD \):
\( \angle BOD = 2\angle BCD \) ...(2)
Adding equation (1) and equation (2):
\( \angle AOC + \angle BOD = 2(\angle ABC + \angle BCD) \) ...(3)
Now examine \( \Delta BCE \):
The exterior angle of a triangle equals the sum of its two interior opposite angles. Side \( BE \) extends along line \( AB \), so \( \angle AEC \) is an exterior angle to \( \Delta BCE \):
\( \angle AEC = \angle EBC + \angle ECB = \angle ABC + \angle BCD \)
Substitute this into equation (3):
\( \angle AOC + \angle BOD = 2\angle AEC \)
Hence proved.
In simple words: The central angles equal twice the angles at B and C. In triangle BCE, the outer angle AEC equals angle B plus angle C, which directly proves the relation.

Exam Tip: Mention the construction "Join BC" at the start and quote the Exterior Angle Theorem for \( \Delta BCE \).

Question 41. If OA and OB are respectively perpendicular to chords CD and EF of a circle whose centre is O. if OA = OB, prove that CD=EF.
Answer: We are given a circle with centre \( O \), \( OA \perp CD \), \( OB \perp EF \), and \( OA = OB \).
Join segments \( OD \) and \( OF \) to form radii of the circle:
\( OD = OF = r \)
In right-angled triangles \( \Delta OAD \) and \( \Delta OBF \):
- \( \angle OAD = \angle OBF = 90^\circ \) (given)
- Hypotenuse \( OD = OF \) (radii of the same circle)
- Side \( OA = OB \) (given)
By the RHS congruence criterion:
\( \Delta OAD \cong \Delta OBF \)
By CPCT:
\( AD = BF \)
Because the perpendicular from the centre to a chord bisects the chord:
\( CD = 2 \times AD \quad \text{and} \quad EF = 2 \times BF \)
Multiplying \( AD = BF \) by 2 gives:
\( 2AD = 2BF \)
\( \implies CD = EF \)
Hence proved.
In simple words: Right triangles OAD and OBF are congruent by RHS, which makes half-chord AD equal to half-chord BF. Doubling them shows that chord CD equals chord EF.

Exam Tip: This question proves Theorem 10.7: "Chords equidistant from the centre of a circle are equal in length." Ensure you mention both the RHS congruence and the chord bisection step.