Maharashtra Board Class 10 Maths Chapter 3 Circle Set 3 Solutions

Question 1. Four alternative answers for each of the following questions are given. Choose the correct alternative.
(i) Two circles of radii 5.5 cm and 3.3 cm respectively touch each other. What is the distance between their centres?
(A) 4.4 cm
(B) 8.8 cm
(C) 2.2 cm
(D) 8.8 or 2.2 cm
Answer: (D) 8.8 or 2.2 cm Two circles can touch each other internally or externally.
\( \therefore \) Distance between centres = 5.5 + 3.3 or 5.5 - 3.3 = 8.8 or 2.2 In simple words: When two circles touch, the distance between their centers is either the sum of their radii (external touch) or the absolute difference of their radii (internal touch).

🎯 Exam Tip: Remember to consider both external and internal touching cases for circles when calculating the distance between their centers. This often leads to two possible answers.

(ii) Two circles intersect each other such that each circle passes through the centre of the other. If the distance between their centres is 12, what is the radius of each circle?
(A) 6 cm
(B) 12 cm
(C) 24 cm
(D) can't say
Answer: (B) 12 cm
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो वृत्तों को दर्शाता है जो एक-दूसरे को प्रतिच्छेद करते हैं। प्रत्येक वृत्त दूसरे वृत्त के केंद्र से होकर गुजरता है। इसमें दो केंद्र P और Q हैं, और एक बिंदु (माना R) है जहाँ वृत्त प्रतिच्छेद करते हैं। PQ दोनों वृत्तों के केंद्रों के बीच की दूरी है, और PR या QR प्रत्येक वृत्त की त्रिज्या है। PQ is the radius = 12 cm In simple words: When two circles intersect and each passes through the other's center, the distance between their centers is equal to the radius of each circle.

🎯 Exam Tip: Visualizing the geometry of intersecting circles is key. If each circle passes through the other's center, the common chord forms two equilateral triangles with the radii, making the distance between centers equal to the radius.

(iii) A circle touches all sides of a parallelogram. So the parallelogram must be a __________
(A) rectangle
(B) rhombus
(C) square
(D) trapezium
Answer: (B) rhombus
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जो एक चतुर्भुज ABCD की चारों भुजाओं को स्पर्श करता है। वृत्त केंद्र में स्थित है और चतुर्भुज की प्रत्येक भुजा स्पर्श बिंदु पर वृत्त की स्पर्श रेखा है। JABCD is a rhombus. Note: It cannot be square as the angles are not mentioned as \( 90^\circ \). In simple words: If a circle can be inscribed within a parallelogram (touching all four sides), then the parallelogram must have all its sides equal in length, which defines a rhombus.

🎯 Exam Tip: A key property to remember is that a quadrilateral circumscribing a circle has the sum of opposite sides equal. For a parallelogram, this condition forces all sides to be equal, making it a rhombus.

(iv) Length of a tangent segment drawn from a point which is at a distance 12.5 cm from the centre of a circle is 12 cm, find the diameter of the circle.
(A) 25 cm
(B) 24 cm
(C) 7 cm
(D) 14 cm
Answer: (C) 7 cm
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र O है। वृत्त के बाहर एक बिंदु A से एक स्पर्श रेखा AB खींची गई है, जिसकी लंबाई 12 cm है। बिंदु A की केंद्र O से दूरी (OA) 12.5 cm है, और B स्पर्श बिंदु है। In \( \triangle OAB \), \( \angle B = 90^\circ \) [Tangent theorem]
\( \therefore OA^2 = OB^2 + AB^2 \) [Pythagoras theorem]
\( \therefore 12.5^2 = OB^2 + 12^2 \)
\( \therefore OB^2 = 156.25 - 144 \)
\( \therefore OB = \sqrt{12.25} = 3.5 \text{ cm} \)
\( \therefore \) Diameter = \( 2 \times OB = 2 \times 3.5 = 7 \text{ cm} \) In simple words: The tangent to a circle is perpendicular to the radius at the point of tangency. This forms a right-angled triangle, allowing us to use the Pythagorean theorem to find the radius and then the diameter.

🎯 Exam Tip: The tangent-radius property (tangent is perpendicular to radius at point of contact) is fundamental. Always look for right-angled triangles to apply the Pythagorean theorem in such problems.

(v) If two circles are touching externally, how many common tangents of them can be drawn?
(A) One
(B) Two
(C) Three
(D) Four
Answer: (C) Three
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो वृत्तों को दर्शाता है जो बाहरी रूप से एक-दूसरे को बिंदु Q पर स्पर्श करते हैं। एक रेखा P और Q के बीच से क्षैतिज रूप से गुजर रही है, और दो अन्य रेखाएं ऊपर और नीचे से वृत्तों को स्पर्श करते हुए गुजर रही हैं। line l, line m and line n are the tangents. In simple words: When two circles touch externally, they have three common tangents: two direct common tangents (above and below) and one transverse common tangent that passes through their point of contact.

🎯 Exam Tip: Visualize the different types of common tangents for circles in various positions (intersecting, touching externally, touching internally, separate). This is a common MCQ question.

(vi) \( \angle ACB \) is inscribed in arc ACB of a circle with centre O. If \( \angle ACB = 65^\circ \), find m(arc ACB).
(A) 65°
(B) 130°
(C) 295°
(D) 230°
Answer: (D) 230°
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र O है। इसमें एक चाप ACB है जिसमें \( \angle ACB \) एक अंतर्लिखित कोण है। बिंदु A और B वृत्त पर हैं, और C वृत्त पर एक बिंदु है जो चाप AB पर है, जिससे कोण ACB बनता है।
\( m \angle ACB = \frac{1}{2} m(\text{arc AB}) \) [Inscribed angle theorem]
\( \therefore m(\text{arc AB}) = 2 m \angle ACB = 2 \times 65 = 130^\circ \)
\( m(\text{arc ACB}) = 360^\circ - m(\text{arc AB}) \) [Measure of a circle is \( 360^\circ \)]
\( = 360^\circ - 130^\circ = 230^\circ \) In simple words: The measure of an inscribed angle is half the measure of its intercepted arc. The measure of the major arc (ACB) is found by subtracting the measure of the minor intercepted arc (AB) from the total \( 360^\circ \) of the circle.

🎯 Exam Tip: Clearly distinguish between the intercepted arc and the arc in which the angle is inscribed. The inscribed angle theorem relates the angle to its intercepted arc. The "arc ACB" usually refers to the major arc containing point C.

(vii) Chords AB and CD of a circle intersect inside the circle at point E. If AE = 5.6, EB = 10, CE = 8, find ED.
(A) 7
(B) 8
(C) 11.2
(D) 9
Answer: (A) 7 Chords AB and CD intersect internally at E.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसमें दो जीवाएँ AB और CD एक दूसरे को वृत्त के भीतर बिंदु E पर काटती हैं। जीवा AB बिंदु A, E, B से बनी है और जीवा CD बिंदु C, E, D से बनी है।
\( AE \times EB = CE \times ED \) [Theorem of internal division of chords]
\( \therefore 5.6 \times 10 = 8 \times ED \)
\( \therefore ED = 7 \) units In simple words: When two chords intersect inside a circle, the product of the segments of one chord is equal to the product of the segments of the other chord.

🎯 Exam Tip: The theorem of internal division of chords \( (AE \times EB = CE \times ED) \) is crucial. Make sure to correctly identify the segments of each chord created by the intersection point.

(viii) In a cyclic \( \Box ABCD \), twice the measure of \( \angle A \) is thrice the measure of \( \angle C \). Find the measure of \( \angle C \)?
(A) 36°
(B) 72°
(C) 90°
(D) 108°
Answer: (B) 72°
\( \angle A + \angle C = 180^\circ \) [Theorem of cyclic quadrilateral]
\( \therefore 2 \angle A + 2 \angle C = 2 \times 180^\circ \) [Multiplying both sides by 2]
\( \therefore 3 \angle C + 2 \angle C = 360^\circ \) [\( \because 2 \angle A = 3 \angle C \)]
\( \therefore 5 \angle C = 360^\circ \)
\( \therefore \angle C = 72^\circ \) In simple words: In a cyclic quadrilateral, opposite angles are supplementary (add up to 180°). By using the given relationship between \( \angle A \) and \( \angle C \), we can solve for \( \angle C \).

🎯 Exam Tip: Recall the properties of cyclic quadrilaterals, especially that opposite angles sum to \( 180^\circ \). Algebraic substitution is then used to solve for the unknown angle.

(ix) Points A, B, C are on a circle, such that m(arc AB) = m(arc BC) = 120°. No point, except point B, is common to the arcs. Which is the type of \( \triangle ABC \)?
(A) Equilateral triangle
(B) Scalene triangle
(C) Right angled triangle
(D) Isosceles triangle
Answer: (A) Equilateral triangle
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिस पर तीन बिंदु A, B और C स्थित हैं। इन बिंदुओं को जोड़कर एक त्रिभुज ABC बनता है। चाप AB, BC और AC त्रिभुज की भुजाओं के संगत हैं।
\( m(\text{arc AB}) + m(\text{arc BC}) + m(\text{arc AC}) = 360^\circ \) [Measure of a circle is \( 360^\circ \)]
\( \therefore 120^\circ + 120^\circ + m(\text{arc AC}) = 360^\circ \)
\( \therefore m(\text{arc AC}) = 120^\circ \)
\( \therefore \text{arc AB} = \text{arc BC} = \text{arc AC} \)
\( \therefore \text{seg AB} \cong \text{seg BC} \cong \text{seg AC} \) [Corresponding chords of congruent arcs of a circle are congruent]
\( \therefore \triangle ABC \) is an equilateral triangle. In simple words: If all three arcs formed by three points on a circle have equal measures, then the chords subtending these arcs are also equal. This means the triangle formed by these points has equal sides, making it an equilateral triangle.

🎯 Exam Tip: Remember that congruent arcs in a circle subtend congruent chords. If the measures of the arcs are equal, the sides of the inscribed triangle will also be equal, leading to an equilateral triangle.

(x) Seg XZ is a diameter of a circle. Point Y lies in its interior. How many of the following statements are true?
(i) It is not possible that \( \angle XYZ \) is an acute angle.
(ii) \( \angle XYZ \) can't be a right angle.
(iii) \( \angle XYZ \) is an obtuse angle.
(iv) Can't make a definite statement for measure of \( \angle XYZ \).
Answer: (C) Only three x. seg XZ is the diameter.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसमें XZ व्यास है और केंद्र वृत्त के मध्य में है। वृत्त पर एक बिंदु W है। वृत्त के अंदर एक बिंदु Y है।
\( \therefore \angle XWZ \) is a right angle. [Angle inscribed in a semicircle] Since, Y lies in the interior of \( \triangle XWZ \),
\( \therefore \angle XYZ > 90^\circ \) i.e., \( \angle XYZ \) is an obtuse angle. In simple words: An angle inscribed in a semicircle is a right angle. If a point Y lies inside the triangle formed by the diameter and an inscribed angle, then the angle at Y subtended by the diameter will be obtuse (greater than 90 degrees).

🎯 Exam Tip: Understand the relationship between an angle inscribed in a semicircle and an angle formed by the diameter with a point inside/outside the semicircle. An angle subtended by the diameter at any point in the interior of the circle is obtuse, at any point on the circle is right, and at any point in the exterior of the circle is acute.

Question 2.

Line I touches a circle with centre O at point P. If radius of the circle is 9 cm, answer the following.
(i) What is d(O, P) = ? Why?
(ii) If d(O, Q) = 8 cm, where does the point Q lie?
(iii) If d(O, R) = 15 cm, how many locations of point R are on line I ? At what distance will each of them be from point P?
Answer:
(i) seg OP is the radius of the circle.
\( \therefore d(O, P) = 9 \text{ cm} \)
(ii) Here, 8 cm < 9 cm
\( \therefore d(O, Q) < d(O, P) \)
\( \therefore d(O, Q) < \text{radius} \) Point Q lies in the interior of the circle.
(iii) There can be two locations of point R on line I.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र O है और एक रेखा 'l' है जो बिंदु P पर वृत्त को स्पर्श करती है। केंद्र O से रेखा 'l' पर दो बिंदु R चिह्नित हैं, जो P के दोनों ओर समान दूरी पर हैं।
\( d(O, R) = 15 \text{ cm} \) Now, in \( \triangle OPR, \angle OPR = 90^\circ \) [Tangent theorem]
\( \therefore OR^2 = OP^2 + PR^2 \) [Pythagoras theorem]
\( \therefore 15^2 = 9^2 + PR^2 \)
\( \therefore 225 = 81 + PR^2 \)
\( \therefore PR^2 = 225 - 81 = 144 \) [Taking square root of both sides]
\( \therefore PR = \sqrt{144} \)
\( = 12 \text{ cm} \) In simple words: The distance from the center to the point of tangency is the radius. Points inside the circle are closer than the radius. For a point on the tangent, the radius at the point of tangency forms a right angle with the tangent, allowing use of the Pythagorean theorem to find distances.

🎯 Exam Tip: The fundamental property that the radius is perpendicular to the tangent at the point of contact is key here. This forms a right-angled triangle, enabling calculations using the Pythagorean theorem for distances involving the center, tangent point, and any point on the tangent.

Question 3.

In the adjoining figure, M is the centre of the circle and seg KL is a tangent segment. If MK = 12, KL = \( 6\sqrt{3} \), then find
(i) Radius of the circle.
(ii) Measures of \( \angle K \) and \( \angle M \).
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र M है। वृत्त के बाहर एक बिंदु K से एक स्पर्श रेखा KL खींची गई है, जहाँ L स्पर्श बिंदु है। रेखाखंड ML वृत्त की त्रिज्या है। Solution: (i) Line KL is the tangent to the circle at point L and seg ML is the radius. [Given]
\( \therefore \angle MLK = 90^\circ \) ............. (i) [Tangent theorem] In \( \triangle MLK, \angle MLK = 90^\circ \)
\( \therefore MK^2 = ML^2 + KL^2 \) [Pythagoras theorem]
\( \therefore 12^2 = ML^2 + (6\sqrt{3})^2 \)
\( \therefore 144 = ML^2 + 108 \)
\( \therefore ML^2 = 144 - 108 \)
\( \therefore ML^2 = 36 \)
\( \therefore ML = \sqrt{36} = 6 \) units. [Taking square root of both sides]
\( \therefore \) Radius of the circle is 6 units.
(ii) We know that,
\( ML = \frac{1}{2} MK \)
\( \therefore \angle K = 30^\circ \) ............... (ii) [Converse of \( 30^\circ - 60^\circ - 90^\circ \) theorem] In \( \triangle MLK \),
\( \angle L = 90^\circ \) [From (i)]
\( \angle K = 30^\circ \) [From (ii)]
\( \therefore \angle M = 60^\circ \) [Remaining angle of \( \triangle MLK \)] In simple words: The radius drawn to the point of tangency is perpendicular to the tangent, forming a right-angled triangle. We use the Pythagorean theorem to find the radius and then properties of a \( 30^\circ - 60^\circ - 90^\circ \) triangle to find the angles.

🎯 Exam Tip: Recognize the formation of a right-angled triangle (radius perpendicular to tangent). If one leg is half the hypotenuse, it's a \( 30^\circ - 60^\circ - 90^\circ \) triangle, simplifying angle calculations.

Question 4.

In the adjoining figure, O is the centre of the circle. Seg AB, seg AC are tangent segments. Radius of the circle is r and l(AB) = r. Prove that, \( \Box ABOC \) is a square.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र O है। वृत्त के बाहर एक बिंदु A से दो स्पर्श रेखाएँ AB और AC खींची गई हैं, जो क्रमशः B और C पर वृत्त को स्पर्श करती हैं। रेखाखंड OB और OC वृत्त की त्रिज्याएँ हैं। Given: O is the centre of circle. seg AB and seg AC are the tangents, radius = r, l(AB) = r. To prove: \( \Box ABOC \) is a square. Construction: Draw seg OB and seg OC. Proof: seg AB and seg AC are the tangents to the circle. [Given]
\( \therefore AB = AC \) [Tangent segment theorem] But, AB = r [Given]
\( \therefore AB = AC = r \) ........... (i) Also, OB = OC = r ........... (ii) [Radii of the same circle]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र केंद्र O वाले वृत्त को दर्शाता है, जिसमें AB और AC स्पर्श रेखाएँ हैं। OB और OC त्रिज्याएँ हैं। यह एक चतुर्भुज ABOC बनाता है।
\( \therefore AB = AC = OB = OC \) [From (i) and (ii)]
\( \therefore \Box ABOC \) is a rhombus.
\( \angle OBA = 90^\circ \) [Tangent theorem]
\( \therefore \Box ABOC \) is a square [A rhombus is a square, if one of its angles is a right angle] In simple words: Since the lengths of tangents from an external point to a circle are equal, and given that the tangent length equals the radius, all four sides of quadrilateral ABOC are equal. Also, the radius is perpendicular to the tangent at the point of contact, making one angle 90 degrees. A rhombus with a right angle is a square.

🎯 Exam Tip: Two key theorems are applied here: the length of tangents from an external point are equal, and the radius is perpendicular to the tangent. Combining these with the definition of a rhombus and a square is essential for the proof.

Question 5.

In the adjoining figure, \( \Box ABCD \) is a parallelogram. It circumscribes the circle with centre T. Points E, F, G, H are touching points. If AE = 4.5, EB = 5.5, find AD.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र T है, और यह एक समांतर चतुर्भुज ABCD की सभी चार भुजाओं को स्पर्श करता है। E, F, G, H क्रमशः भुजाओं AB, BC, CD, DA पर स्पर्श बिंदु हैं। Solution: Let the values of DH and CF be x and y respectively. [ \( AE = AH = 4.5 \) \( BE = BF = 5.5 \) \( DH = DG = x \) \( CF = CG = y \) ] [Tangent segment theorem] \( \Box ABCD \) is a parallelogram. [Given]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र ऊपर दिए गए समांतर चतुर्भुज और वृत्त को फिर से दर्शाता है, जिसमें स्पर्श बिंदुओं पर खंडों की लंबाई (AE=4.5, EB=5.5, AH=4.5, BF=5.5) और अज्ञात लंबाई x और y (DG=x, DH=x, CG=y, CF=y) चिह्नित हैं।
\( \therefore AB = CD \) [Opposite sides of a parallelogram]
\( \therefore AE + BE = DG + CG \) [A - E - B, D - G - C]
\( \therefore 4.5 + 5.5 = x + y \)
\( \therefore x + y = 10 \) ........... (i) Also, AD = BC [Opposite sides of a parallelogram]
\( \therefore AH + DH = BF + CF \) [A - H - D, B - F - C]
\( \therefore 4.5 + x = 5.5 + y \)
\( \therefore x - y = 1 \) ............ (ii) Adding equations (i) and (ii), we get
\( 2x = 11 \)
\( \therefore x = \frac{11}{2} = 5.5 \)
\( \therefore AD = AH + DH \) [A - H - D]
\( = 4.5 + 5.5 \)
\( \therefore AD = 10 \) units In simple words: In a quadrilateral circumscribing a circle, the lengths of tangents from a vertex to the circle are equal. For a parallelogram, opposite sides are equal. Using these properties, we can set up and solve a system of equations to find the unknown side length.

🎯 Exam Tip: Remember the property that tangents from an external point to a circle are equal. Combine this with the properties of a parallelogram (opposite sides are equal) to form equations and solve for unknown lengths.

Question 6.

In the adjoining figure, circle with centre M touches the circle with centre N at point T. Radius RM touches the smaller circle at S. Radii of circles are 9 cm and 2.5 cm. Find the answers to the following questions, hence find the ratio MS : SR.
(i) Find the length of segment MT.
(ii) Find the length of seg MN.
(iii) Find the measure of \( \angle NSM \).
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो वृत्तों को दर्शाता है। बड़ा वृत्त जिसका केंद्र M है और छोटा वृत्त जिसका केंद्र N है, बिंदु T पर एक दूसरे को स्पर्श करते हैं। एक रेखा RM बड़े वृत्त की त्रिज्या है और यह छोटे वृत्त को बिंदु S पर स्पर्श करती है। Solution: (i) MT = 9 cm [Radius of the bigger circle]
(ii) MT = MN + NT [M - N - T]
\( \therefore 9 = MN + 2.5 \)
\( \therefore MN = 9 - 2.5 \)
\( \therefore MN = 6.5 \text{ cm} \)
(iii) seg MR is a tangent to the smaller circle and NS is its radius.
\( \therefore \angle NSM = 90^\circ \) [Tangent theorem] In \( \triangle NSM, \angle NSM = 90^\circ \)
\( \therefore MN^2 = NS^2 + MS^2 \) [Pythagoras theorem]
\( \therefore 6.5^2 = 2.5^2 + MS^2 \)
\( \therefore MS^2 = 6.5^2 - 2.5^2 \)
\( = (6.5 + 2.5) (6.5 - 2.5) \) [\( \because a^2 - b^2 = (a + b)(a - b) \)]
\( = 9 \times 4 = 36 \)
\( \therefore MS = \sqrt{36} \) [Taking square root of both sides]
\( = 6 \text{ cm} \) But, MR = MS + SR [M - S - R]
\( \therefore 9 = 6 + SR \)
\( \therefore SR = 9 - 6 \)
\( \therefore SR = 3 \text{cm} \) Now, \( \frac{MS}{SR} = \frac{6}{3} = \frac{2}{1} \)
\( \therefore MS : SR = 2 : 1 \) In simple words: The distance between the centers of two externally touching circles is the sum of their radii. The radius of a circle is perpendicular to the tangent at the point of contact, forming a right-angled triangle, which helps in calculating unknown lengths using the Pythagorean theorem and then finding the required ratio.

🎯 Exam Tip: Break down the problem into smaller geometric principles: touching circles (centers and tangency point are collinear), tangent-radius perpendicularity (forms a right triangle), and the Pythagorean theorem. Pay attention to segment addition postulates.

Question 7.

In the adjoining figure, circles with centres X and Y touch each other at point Z. A secant passing through Z intersects the circles at points A and B respectively. Prove that, radius XA || radius YB. Fill in the blanks and complete the proof.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो वृत्तों को दर्शाता है जो एक-दूसरे को बाहरी रूप से बिंदु Z पर स्पर्श करते हैं। केंद्र X वाले वृत्त को केंद्र Y वाले वृत्त से छोटा दिखाया गया है। एक तिर्यक रेखा Z से गुजरती हुई पहले वृत्त को A पर और दूसरे वृत्त को B पर काटती है। Given: X and Y are the centres of circle. To prove: radius XA || radius YB Construction: Draw segments XZ and YZ. Proof: By theorem of touching circles, points X, Z, Y are collinear.
\( \therefore \angle XZA \cong \angle BZY \) [Vertically opposite angles] Let \( \angle XZA = \angle BZY = a \) ............... (i) Now, seg XA \( \cong \) seg XZ [Radii of the same circle]
\( \therefore \angle XAZ = \angle XZA = a \) ............... (ii) [Isosceles triangle theorem] Similarly, seg YB \( \cong \) seg YZ [Radii of the same circle]
\( \therefore \angle BZY = \angle ZBY = a \) ............... (iii) [Isosceles triangle theorem]
\( \therefore \angle XAZ = \angle ZBY \) [From (i), (ii) and (iii)]
\( \therefore \) radius XA || radius YB [Alternate angles test] In simple words: When two circles touch, their centers and the point of contact are collinear. By using the isosceles triangle theorem for radii and vertically opposite angles, we can show that corresponding angles formed by the secant and radii are equal, proving the radii are parallel.

🎯 Exam Tip: The collinearity of centers and the point of tangency is a critical starting point. Then, identifying isosceles triangles formed by radii and using angle properties (like vertically opposite or alternate angles) are key steps to complete such proofs.

Question 8.

In the adjoining figure, circles with centres X and Y touch internally at point Z. Seg BZ is a chord of bigger circle and it intersects smaller circle at point A. Prove that, seg AX || seg BY.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो वृत्तों को दर्शाता है जो आंतरिक रूप से बिंदु Z पर स्पर्श करते हैं। बड़ा वृत्त जिसका केंद्र X है और छोटा वृत्त जिसका केंद्र Y है। बड़े वृत्त की जीवा BZ छोटे वृत्त को बिंदु A पर काटती है। Given: X and Y are the centres of the circle. To prove: seg AX || seg BY Proof: In \( \triangle XAZ \), seg XA \( \cong \) seg XZ [Radii of the same circle]
\( \therefore \angle XZA = \angle XAZ \) ............ (i) [Isosceles triangle theorem] Also, in \( \triangle YBZ \), seg YB \( \cong \) seg YZ [Radii of the sanie circle]
\( \therefore \angle YZB = \angle YBZ \) [Isosceles triangle theorem]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र आंतरिक रूप से स्पर्श करने वाले वृत्तों को फिर से दर्शाता है, जिसमें XZ और YZ रेखाखंडों को दर्शाया गया है जो क्रमशः केंद्र X और Y से बिंदु Z तक हैं। BZ जीवा, A से गुजर रही है।
\( \therefore \angle XZA \cong \angle YBZ \) ............... (ii) [Y - X - Z, B - A - Z]
\( \therefore \angle XAZ \cong \angle YBZ \) [From (i) and (ii)]
\( \therefore \) seg AX || seg BY [Corresponding angles test] In simple words: For internally touching circles, their centers and the point of contact are collinear. By forming isosceles triangles with radii and using the property of corresponding angles, we can prove that the segments AX and BY are parallel.

🎯 Exam Tip: The principle of collinearity of centers and the point of internal tangency is crucial. Recognize isosceles triangles formed by radii and use angle relationships to prove parallelism, specifically corresponding angles.

Question 9.

In the adjoining figure, line I touches the circle with centre O at point P, Q is the midpoint of radius OP. RS is a chord through Q such that chords RS || line I. If RS = 12, find the radius of the circle.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र O है। एक रेखा 'l' बिंदु P पर वृत्त को स्पर्श करती है। त्रिज्या OP का मध्यबिंदु Q है। एक जीवा RS बिंदु Q से होकर गुजरती है और रेखा 'l' के समानांतर है। Solution: Let the radius of the circle be r. line l is the tangent to the circle and [Given] seg OP is the radius.
\( \therefore \) seg OP \( \perp \) line l [Tangent theorem] chord RS || line l [Given]
\( \therefore \) seg OP \( \perp \) chord RS
\( \therefore QS = \frac{1}{2} RS \) [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
\( = \frac{1}{2} \times 12 = 6 \text{ cm} \) Also, \( OQ = \frac{1}{2} OP \) [Q is the midpoint of OP]
\( = \frac{1}{2} r \) In \( \triangle OQS, \angle OQS = 90^\circ \) [seg OP \( \perp \) chord RS ]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र ऊपर दिए गए वृत्त को फिर से दर्शाता है, जिसमें केंद्र O, स्पर्श बिंदु P, रेखा 'l', त्रिज्या OP का मध्यबिंदु Q, और जीवा RS (रेखा 'l' के समानांतर) दर्शाए गए हैं।
\( \therefore OS^2 = OQ^2 + QS^2 \) [Pythagoras theorem]
\( \therefore r^2 = (\frac{1}{2}r)^2 + 6^2 \)
\( \therefore r^2 = \frac{1}{4} r^2 + 36 \)
\( \therefore r^2 - \frac{1}{4} r^2 = 36 \)
\( \therefore \frac{3}{4} r^2 = 36 \)
\( \therefore r^2 = \frac{36 \times 4}{3} \)
\( \therefore r^2 = 48 \)
\( \therefore r = \sqrt{48} \) [Taking square root of both sides]
\( = 4\sqrt{3} \)
\( \therefore \) The radius of the given circle is \( 4\sqrt{3} \text{ cm} \). In simple words: We use the properties that the radius is perpendicular to the tangent, and a line perpendicular from the center bisects a chord. These create a right-angled triangle, allowing us to find the radius using the Pythagorean theorem.

🎯 Exam Tip: This problem integrates several theorems: tangent-radius perpendicularity, perpendicular from center to chord bisects it, and properties of parallel lines. Carefully apply each to form the right-angled triangle for the Pythagorean theorem.

Question 10.

In the adjoining figure, seg AB is a diameter of a circle with centre C. Line PQ is a tangent, which touches the circle at point T. Seg AP \( \perp \) line PQ and seg BQ \( \perp \) line PQ. Prove that seg CP \( \cong \) seg CQ.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र C है और AB उसका व्यास है। एक रेखा PQ वृत्त को बिंदु T पर स्पर्श करती है। बिंदु A और B से रेखा PQ पर क्रमशः AP और BQ लंब खींचे गए हैं। Given: C is the centre of circle. seg AB is the diameter of circle. line PQ is a tangent, seg AP \( \perp \) line PQ and seg BQ \( \perp \) line PQ. To prove: seg CP \( \cong \) seg CQ Construction: Draw seg CT, seg CP and seg CQ. Proof: Line PQ is the tangent to the circle at point T. [Given]
\( \therefore \) seg CT \( \perp \) line PQ (i) [Tangent theorem] Also, seg AP \( \perp \) line PQ, seg BQ \( \perp \) line PQ [Given]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र ऊपर दिए गए वृत्त को फिर से दर्शाता है, जिसमें केंद्र C से T तक त्रिज्या CT खींची गई है। AP, CT और BQ सभी रेखा PQ पर लंबवत हैं।
\( \therefore \) seg AP || seg CT || seg BQ [Lines perpendicular to the same line are parallel to each other]
\( \frac{AC}{CB} = \frac{PT}{TQ} \) [Property of intercepts made by three parallel lines and their transversals] But, AC = CB [Radii of the same circle]
\( \therefore \frac{AC}{CB} = \frac{PT}{TQ} = 1 \)
\( \therefore PT = TQ \) ............ (ii) Now, in \( \triangle CTP \) and \( \triangle CTQ \), seg PT \( \cong \) seg QT [From (ii)] \( \angle CTP = \angle CTQ \) [From (i), each angle is of measure \( 90^\circ \)] seg CT \( \cong \) seg CT [Common side]
\( \therefore \triangle CTP \cong \triangle CTQ \) [SAS test of congruence]
\( \therefore \) seg CP \( \cong \) seg CQ [c.s.c.t] In simple words: By proving that AP, CT, and BQ are parallel lines, we can apply the intercept theorem to show that T is the midpoint of PQ. Then, using the tangent-radius perpendicularity and SAS congruence criterion, we can prove that \( \triangle CTP \) is congruent to \( \triangle CTQ \), leading to CP being congruent to CQ.

🎯 Exam Tip: This proof combines parallelism, the intercept theorem, tangent-radius property, and triangle congruence. Clearly state each theorem and its application. The SAS congruence for \( \triangle CTP \) and \( \triangle CTQ \) is the final step to prove segment congruence.

Question 11. Draw circles with centres A, B and C each of radius 3 cm, such that each circle touches the other two circles. Analysis: Let the circles with centres A, B, C touch each other at points P, Q, R. [.: A - P - B A-Q-C B – R – C ] [Theorem of touching circles]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र तीन वृत्तों को दर्शाता है जिनके केंद्र A, B और C हैं। प्रत्येक वृत्त की त्रिज्या 3 सेमी है। तीनों वृत्त एक-दूसरे को बाहरी रूप से स्पर्श कर रहे हैं, जिससे उनके केंद्रों A, B, C को जोड़ने पर एक समबाहु त्रिभुज बनता है। स्पर्श बिंदु P, Q, R हैं।
Answer:
\( \therefore \) AB = AP + BP [A – P – B]
\( \therefore \) AB = 3 + 3 = 6 cm Similarly, BC = 6 cm, AC = 6 cm So, if we construct triangle \(\triangle\)ABC of side 6 cm each, then with A, B, C as the centres In simple words: To draw three circles each with a 3 cm radius touching each other, we form an equilateral triangle with their centers, where each side length is twice the radius (6 cm).

🎯 Exam Tip: Understanding the relationship between radii and distances between centers for touching circles is crucial for scoring well in geometry problems.

Question 12. Prove that any three points on a circle cannot be collinear Given: A circle with centre O. Points A, B and C lie on the circle. To prove: Points A, B and C are not collinear. Proof: OA = OB [Radii of the same circle]
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक वृत्त है जिसका केंद्र O है। वृत्त पर तीन बिंदु A, B और C स्थित हैं। केंद्र O से इन बिंदुओं तक त्रिज्याएँ OA, OB और OC खींची गई हैं। इसके अलावा, AB और BC को जोड़ने वाले दो खंड भी दिखाए गए हैं, और इन खंडों के लंब समद्विभाजक (m और l) बिंदु O से गुजरते हुए दर्शाए गए हैं।
Answer:
\( \therefore \) Point O is equidistant from the endpoints A and B of seg AB.
\( \therefore \) Point O lies on the perpendicular bisector of AB. [Perpendicular bisector theorem] Similarly, we can prove that, Point O lies on the perpendicular bisector of BC.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त के अंदर के ज्यामितीय संबंध को दर्शाता है। इसमें तीन बिंदु A, B और C हैं। केंद्र O को दिखाया गया है। रेखाएं L और M क्रमशः AB और BC के लंब समद्विभाजक हैं, जो बिंदु O पर मिलती हैं। यह दर्शाता है कि O, \(\triangle\)ABC का परिकेंद्र है।
Answer:
\( \therefore \) Point O is the point of intersection of perpendicular bisectors of AB and BC (i.e., circumcentre of \(\triangle\)ABC) ............. (i) Now, suppose that the points A, B, C are collinear. Then, the perpendicular bisector of AB and BC will be parallel. [Perpendiculars to the same line are parallel]
\( \therefore \) The perpendicular bisector do not intersect at O. This contradicts statement (i) that the perpendicular bisectors intersect each other at O.
\( \therefore \) Our supposition that A, B, C are collinear is false.
\( \therefore \) Points A, B and C are non collinear points. In simple words: If three points on a circle were collinear, their perpendicular bisectors would be parallel and never meet, which contradicts the fact that the circle's center (circumcenter) is where they must intersect. Hence, points on a circle cannot be collinear.

🎯 Exam Tip: This is a proof by contradiction. Clearly stating the initial assumption and showing how it leads to a contradiction is key to a complete answer.

Question 13. In the adjoining figure, line PR touches the circle at point Q. Answer the following questions with the help of the figure.
(i) What is the sum of \(\angle\)TAQ and \(\angle\)TSQ?
(ii) Find the angles which are congruent to \(\angle\)AQP.
(iii) Which angles are congruent to \(\angle\)QTS?
(iv) \(\angle\)TAS = 65°, find the measures of \(\angle\)TQS and arc TS.
(v) If \(\angle\)AQP = 42° and \(\angle\)SQR = 58° find measure of \(\angle\)ATS.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसमें एक स्पर्शरेखा PR बिंदु Q पर वृत्त को स्पर्श कर रही है। वृत्त के अंदर एक चक्रीय चतुर्भुज AQST बना हुआ है। इसमें विभिन्न जीवाएँ और चाप दिखाए गए हैं, और स्पर्शरेखा के साथ बने कोणों का अध्ययन किया जा रहा है।
Answer: i. AQST is a cyclic quadrilateral. [Given]
\( \therefore \) \(\angle\)TAQ + \(\angle\)TSQ = 180° [Opposite angles of a cyclic quadrilateral are supplementary] ii. line PR is the tangent and seg AQ is the secant. [Given]
\( \therefore \) \(\angle\)AQP = \( \frac{1}{2} \) m(arc AQ) [Theorem of angle between tangent and secant] But, \(\angle\)ASQ = \( \frac{1}{2} \) m(arc AQ) [Inscribed angle theorem]
\( \therefore \) \(\angle\)AQP \( \cong \) \(\angle\)ASQ Similarly, we can prove that, \(\angle\)AQP = \(\angle\)ATQ iii. \(\angle\)QTS = \( \frac{1}{2} \) m(arc QS) [Inscribed angle theorem] But, \(\angle\)SQR = \( \frac{1}{2} \) m(arc QS) [Theorem of angle between tangent and secant]
\( \therefore \) \(\angle\)QTS \( \cong \) \(\angle\)SQR Also, \(\angle\)QTS = \(\angle\)QAS [Angles inscribed in the same arc] iv. \(\angle\)TQS = \(\angle\)TAS [Angles inscribed in the same arc]
\( \therefore \) \(\angle\)TQS = 65° Now, \(\angle\)TQS = \( \frac{1}{2} \) m(arc TS) [Inscribed angle theorem]
\( \therefore \) 65°= \( \frac{1}{2} \) m(arcTS)
\( \therefore \) m(arc TS) = 65° \( \times \) 2
\( \therefore \) m(arc TS) = 130° v. \(\angle\)AQP + \(\angle\)AQS + \(\angle\)SQR = 180° [Angles in a linear pair]
\( \therefore \) 42° + \(\angle\)AQS + 58° = 180°
\( \therefore \) \(\angle\)AQS + 100° = 180° ............. (i) But, AQST is a cyclic quadrilateral.
\( \therefore \) \(\angle\)AQS + \(\angle\)ATS = 180° ............ (ii) [Theorem of cyclic quadrilateral]
\( \therefore \) \(\angle\)ATS = 100° [From (i) and (ii)] In simple words: This question tests understanding of angles in cyclic quadrilaterals, angles formed by tangents and secants, and inscribed angles. Key concepts involve supplementary opposite angles in a cyclic quadrilateral and the relationship between an inscribed angle and its intercepted arc.

🎯 Exam Tip: For problems involving tangents, secants, and cyclic quadrilaterals, carefully identify the intercepted arcs and apply the correct theorems. Listing given information and reasons systematically is crucial.

Question 14. In the adjoining figure, O is the centre of a circle, chord PQ \( \cong \) chord RS. If \(\angle\)POR = 70° and (arc RS) = 80°, find
(i) m (arc PR)
(ii) m (arc QS)
(iii) m (arc QSR).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र O है। इसमें दो जीवाएँ PQ और RS हैं। बिंदु P, R, Q, S वृत्त की परिधि पर स्थित हैं। केंद्र O से P और R तक रेखाखंड (त्रिज्या) खींचे गए हैं, जिससे \(\angle\)POR बनता है।
Answer: Solution: i. m(arc PR) = m\(\angle\)POR [Definition of measure of arc]
\( \therefore \) m(arc PR) = 70° ii. chord PQ \( \cong \) chord RS [Given]
\( \therefore \) m(arc PQ) = m(arc RS) = 80° [Corresponding arcs of congruents chords of a circle are congruent] Now, m(arc QS) + m(arc PQ) + m(arc PR) + m(arcRS) = 360°
\( \therefore \) m(arc QS) + 80° + 70° + 80° = 360° [Measure of a circle is 360°]
\( \therefore \) m(arc QS) + 230° = 360°
\( \therefore \) m(arc QS) = 130° iii. m(arc QSR) = m(arc QS) + m(arc SR) [Arc addition property] = 130° + 80°
\( \therefore \) m(arc QSR) = 210° In simple words: This problem involves finding arc measures using central angles, congruent chords, and the total measure of a circle. Congruent chords subtend congruent arcs, and the measure of a central angle equals the measure of its intercepted arc.

🎯 Exam Tip: Remember that the entire circle measures 360°. When given congruent chords, their corresponding arcs are equal in measure. Break down complex arcs into sums of simpler ones.

Question 15. In the adjoining figure, m(arc WY) = 44°, m(arc ZX) = 68°, then
(i) Find the measure of \(\angle\)ZTX.
(ii) If WT = 4.8, TX = 8.0, YT = 6.4, find TZ.
(iii) If WX = 25, YT = 8, YZ = 26, find WT.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसमें दो जीवाएँ WX और YZ बिंदु T पर आंतरिक रूप से प्रतिच्छेद कर रही हैं। विभिन्न चापों (WY, ZX) को भी दर्शाया गया है। यह आकृति जीवाओं के आंतरिक प्रतिच्छेदन से बनने वाले कोणों और खंडों के गुणों का वर्णन करती है।
Answer: i. Chords WX and YZ intersect internally at point T.
\( \therefore \) \(\angle\)ZTX = \( \frac{1}{2} \) [m(arc WY) + m(arc ZX)] = \( \frac{1}{2} \) (44° + 68°) = \( \frac{1}{2} \) \( \times \) 112°
\( \therefore \) m \(\angle\)ZTX = 56° ii. WT \( \times \) TX = YT \( \times \) TZ [Theorem of internal division of chords]
\( \therefore \) 4.8 \( \times \) 8.0 = 6.4 \( \times \) TZ
\( \therefore \) TZ = \( \frac{4.8 \times 8.0}{6.4} \)
\( \therefore \) (TZ) = 6.0 units iii. Let the value of WT be x. [W – T – X] WT + TX = WX
\( \therefore \) x + TX = 25
\( \therefore \) TX = 25 - x Also, YT + TZ = YZ [Y – T – Z]
\( \therefore \) 8 + TZ = 26
\( \therefore \) TZ = 26 - 8 = 18 units But, WT \( \times \) TX = YT \( \times \) TZ [Theorem of internal division of chords]
\( \therefore \) x \( \times \) (25 - x) = 8 \( \times \) 18
\( \therefore \) 25x - x\(^{2}\) = 144
\( \therefore \) x\(^{2}\) - 25x + 144 = 0
\( \therefore \) (x - 16)(x - 9) = 0
\( \therefore \) x = 16 or x = 9
\( \therefore \) WT = 16 units or WT = 9units In simple words: The theorem of internal division of chords states that for two intersecting chords, the product of the segments of one chord equals the product of the segments of the other chord. This principle is used to find unknown lengths.

🎯 Exam Tip: For problems involving segments of intersecting chords, set up the product equation correctly. If finding a segment length from a total length, use variable substitution (e.g., x and total-x).

Question 16. In the adjoining figure,
(i) m (arc CE) = 54°, m (arc BD) = 23°, find measure of \(\angle\)CAE.
(ii) If AB = 4.2,BC = 5.4, AE = 12.0, find AD.
(iii) If AB = 3.6, AC = 9.0, AD = 5.4, find AE.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त और एक छेदक रेखा को दर्शाता है। वृत्त में जीवाएँ BC और ED हैं जो वृत्त के बाहर बिंदु A पर प्रतिच्छेद करती हैं। यह आकृति जीवाओं के बाह्य प्रतिच्छेदन से बने कोणों और खंडों के गुणों का वर्णन करती है।
Answer: i. Chords BC and ED intersect each other externally at point A.
\( \therefore \) \(\angle\)CAE = \( \frac{1}{2} \) [m(arc CE) - m(arc BD)] = \( \frac{1}{2} \) (54° - 23°) = \( \frac{1}{2} \) \( \times \) 31°
\( \therefore \) m\(\angle\)CAE = 15.5° ii. AC = AB + BC [A - B - C] = 4.2 + 5.4 = 9.6 units Now, AB \( \times \) AC = AD \( \times \) AE [Theorem of external division of chords]
\( \therefore \) 4.2 \( \times \) 9.6 = AD \( \times \) 12.0
\( \therefore \) AD = \( \frac{4.2 \times 9.6}{12.0} \)
\( \therefore \) AD = 3.36 units iii. AB \( \times \) AC = AD \( \times \) AE [Theorem of external division of chords]
\( \therefore \) 3.6 \( \times \) 9.0 = 5.4 \( \times \) AE
\( \therefore \) AE = \( \frac{3.6 \times 9.0}{5.4} \)
\( \therefore \) AE = 6 units In simple words: When two secants intersect outside a circle, the measure of the angle formed is half the difference of the measures of the intercepted arcs. Also, the product of the length of the entire secant segment and its external part is constant. These theorems are used to find unknown angles and segment lengths.

🎯 Exam Tip: Distinguish between internal and external intersection of chords/secants. For external intersection, the angle is half the *difference* of the intercepted arcs, and for segments, use the product of the whole secant and its external part.

Question 17. In the adjoining figure, chord EF || chord GH. Prove that, chord EG \( \cong \) chord FH. Fill in the blanks and write the proof. Given: chord EF || chord GH To prove: chord EG = chord FH Construction: Draw seg GF.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसमें दो समांतर जीवाएँ EF और GH हैं। वृत्त पर बिंदु E, F, G, H स्थित हैं। जीवा GF को जोड़ा गया है। यह आकृति समांतर जीवाओं द्वारा अंतर्निहित चापों और उनके द्वारा बनाए गए कोणों के गुणों का वर्णन करती है।
Answer: Proof:
\( \angle \)EFG = \(\angle\)FGH (i) Alternate angles
\( \angle \)EFG = \( \frac{1}{2} \) m (arcEG)] (ii) [Inscribed angle theorem]
\( \angle \)FGH = \( \frac{1}{2} \) m(arcFH) (iii) [Inscribed angle theorem]
\( \therefore \) m(arcEG) = m (arc FH) [From (i), (ii) and (iii)]
\( \therefore \) chord EG \( \cong \) chord FH The chords corresponding to congruent arcs of a circle are congruent In simple words: If two chords in a circle are parallel, then the arcs intercepted between them are congruent. Consequently, the chords subtending these congruent arcs are also congruent.

🎯 Exam Tip: The property that parallel chords intercept congruent arcs is fundamental. Use inscribed angle theorem to relate angles to arcs and then chord congruence to arc congruence.

Question 18. In the adjoining figure, P is the point of contact.
(i) If m (arc PR) = 140°, \(\angle\) POR = 36°, find m (arc PQ)
(ii) If OP = 7.2, OQ = 3.2, find OR and QR
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र O है। बिंदु P वृत्त पर स्पर्श बिंदु है। एक छेदक रेखा QOR वृत्त को Q और R पर प्रतिच्छेद करती है, जबकि यह P से बाहर निकलती है। \(\angle\)POR एक केंद्रीय कोण है और चाप PR को दर्शाता है।
Answer: Solution: i. \(\angle\)PQR \( = \frac{1}{2} \) m(arc PR) [Inscribed angle theorem] = \( \frac{1}{2} \) \( \times \) 140° = 70° \(\angle\)PQR is the exterior angle of \(\triangle\)POQ. [Remote interior angle theorem]
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक वृत्त है जिसका केंद्र O है। एक रेखा QOR वृत्त को Q और R पर प्रतिच्छेद करती है और यह बिंदु P पर वृत्त की स्पर्शरेखा भी है। यह आकृति स्पर्शरेखा-छेदक खंडों के प्रमेय को दर्शाती है।
Answer:
\( \therefore \) \(\angle\)PQR = \(\angle\)POQ + \(\angle\)QPO [R - Q - O]
\( \therefore \) 70° = \(\angle\)POR + \(\angle\)QPO
\( \therefore \) 70 = 36° + \(\angle\)QPO
\( \therefore \) \(\angle\)QPO = 70° - 36° = 34° Now, ray OP is tangent at point P and segment PQ is a secant.
\( \therefore \) \(\angle\)QPO = \( \frac{1}{2} \) m(arcPQ) [Theorem of angle between tangent and secant]
\( \therefore \) 34° = \( \frac{1}{2} \) m(arc PQ)
\( \therefore \) m(arc PQ) = 68° ii. Here, OP = 7.2, OQ = 3.2 Line OP is the tangent at point P [Given] and seg OR is the secant.
\( \therefore \) OP\(^{2}\) = OQ \( \times \) OR [Tangent secant segments theorem]
\( \therefore \) 7.2\(^{2}\) = 3.2 \( \times \) OR
\( \therefore \) 51.84 = 3.2 \( \times \) OR
\( \therefore \) OR = \( \frac{51.84}{3.2} \)
\( \therefore \) OR = 16.2 units Now, OR = OQ + QR [O - Q - R]
\( \therefore \) 16.2 = 3.2 + QR
\( \therefore \) QR = 16.2 - 3.2
\( \therefore \) QR = 13 units iii. Here, OP = 7.2, OR = 16.2 OP\(^{2}\) = OQ \( \times \) OR [Tangent secant segments theorem]
\( \therefore \) 7.2\(^{2}\) = OQ \( \times \) 16.2
\( \therefore \) OQ = \( \frac{51.84}{16.2} \)
\( \therefore \) OQ = 3.2 units Now, OR = OQ + QR [O - Q - R]
\( \therefore \) 16.2 = 3.2 + QR
\( \therefore \) QR = 16.2 - 3.2
\( \therefore \) QR = 13 units In simple words: This problem utilizes the inscribed angle theorem, the remote interior angle theorem, and the tangent-secant segment theorem. The tangent-secant theorem states that the square of the tangent segment equals the product of the whole secant segment and its external part, which is key for finding unknown lengths.

🎯 Exam Tip: For problems involving tangents and secants from an external point, ensure you correctly apply the tangent-secant segment theorem (tangent\(^2\) = whole secant \( \times \) external part). This is a common formula for scoring marks.

Question 19. In the adjoining figure, circles with centres C and D touch internally at point E. D lies on the inner circle. Chord EB of the outer circle intersects inner circle at point A. Prove that, seg EA \( \cong \) seg AB.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो वृत्तों को दर्शाता है जो बिंदु E पर आंतरिक रूप से स्पर्श करते हैं। बड़े वृत्त का केंद्र C है और छोटे वृत्त का केंद्र D है। बड़े वृत्त की जीवा EB छोटे वृत्त को बिंदु A पर काटती है। केंद्र D छोटे वृत्त की परिधि पर स्थित है।
Answer: Given: Circles with centres C and D touch each other internally. To prove: seg EA \( \cong \) seg AB Construction: Join seg ED and seg DA. Proof: E - C - D [Theorem of touching circles] seg ED is the diameter of smaller circle.
\( \therefore \) \(\angle\)EAD = 90° [Angle inscribed in a semicircle]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो आंतरिक रूप से स्पर्श करने वाले वृत्तों को दर्शाता है। बड़े वृत्त का केंद्र C है और छोटे वृत्त का केंद्र D है, जो छोटे वृत्त के व्यास ED पर स्थित है। जीवा EB बड़े वृत्त को स्पर्श करते हुए छोटे वृत्त को बिंदु A पर काटती है। रेखाखंड DA जीवा EB पर लंबवत है।
Answer:
\( \therefore \) seg AD \( \perp \) chord EB
\( \therefore \) seg EA \( \cong \) seg AB [Perpendicular drawn from the centre of the circle to the chord bisects the chord] In simple words: When two circles touch internally, their centers and the point of contact are collinear. If a chord in the larger circle becomes a diameter of the smaller circle, the angle inscribed in the smaller circle by that diameter is 90 degrees. A perpendicular from the center to a chord bisects the chord.

🎯 Exam Tip: For problems involving touching circles, remember that their centers and the point of contact are always collinear. The angle inscribed in a semicircle is always a right angle.

Question 20. In the adjoining figure, seg AB is a diameter of a circle with centre O. The bisector of \(\angle\)ACB intersects the circle at point D. Prove that, seg AD = seg BD. Complete the following proof by filling the blanks.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में केंद्र O वाले एक वृत्त का व्यास AB है। बिंदु C वृत्त पर स्थित है, जिससे \(\triangle\)ACB एक अर्धवृत्त में बना कोण है। \(\angle\)ACB का समद्विभाजक CD वृत्त को बिंदु D पर प्रतिच्छेद करता है।
Answer: Given: seg AB is a diameter, seg CD bisects \(\angle\)ACB. To prove: seg AD = seg BD Construction: Draw seg OD. Proof: \(\angle\)ACB = 90° [Angle inscribed in a semicircle] \(\angle\)DCB = \(\angle\)DCA = 45° [CD is the bisector of \(\angle\)C] m(arcDB) = 2\(\angle\)DCA = 90° [Inscribed angle theorem] \(\angle\)DOB = m(arc DB) = 90° ............. (i) [Definition of measure of arc] segOA = segOB ............. (ii) [Radii of the same circle]
\( \therefore \) line OD is the perpendicular bisector of [From (i) and (ii)] seg AB.
\( \therefore \) seg AD \( \cong \) seg BD In simple words: An angle inscribed in a semicircle is 90 degrees. If a line bisects this 90-degree angle and passes through the circle, it creates an arc whose measure is twice the bisected angle. If this arc's central angle is 90 degrees, and the line also connects to the center and passes through the diameter, it implies the line is a perpendicular bisector, making the chord segments equal.

🎯 Exam Tip: Carefully apply the inscribed angle theorem and the definition of arc measure. When a segment from the center is perpendicular to a chord, it bisects the chord, which is a key property for proving segment congruence.

Question 21. In the adjoining figure, seg MN is a chord of a circle with centre O. MN = 25, L is a point on chord MN such that ML = 9 and d(O, L) = 5. Find the radius of the circle.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र O है। MN वृत्त की एक जीवा है। बिंदु L जीवा MN पर इस प्रकार स्थित है कि ML = 9। केंद्र O से L तक की दूरी 5 है। यह आकृति वृत्त की त्रिज्या ज्ञात करने के लिए पाइथागोरस प्रमेय का उपयोग करने को दर्शाती है।
Answer: Construction: Draw seg OK \( \perp \) chord MN. Join OM. Solution: seg OK \( \perp \) chord MN [Construction]
\( \therefore \) MK = \( \frac{1}{2} \) MN [Perpendicular drawn from the centre of the circle to the chord bisects the chord] = \( \frac{1}{2} \) \( \times \) 25 = 12.5 units MK = ML + LK [M - L - K]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र केंद्र O वाले वृत्त में जीवा MN को दिखाता है। OK, MN पर लंबवत है, जहाँ K, MN का मध्यबिंदु है। बिंदु L जीवा MN पर स्थित है। OM वृत्त की त्रिज्या है। यह समकोण त्रिभुज \(\triangle\)OKL और \(\triangle\)OKM को दर्शाता है, जिनका उपयोग त्रिज्या ज्ञात करने के लिए किया जाता है।
Answer:
\( \therefore \) 12.5 = 9 + LK
\( \therefore \) LK= 12.5 - 9 = 3.5 units In \(\triangle\)OKL, \(\angle\)OKL = 90°
\( \therefore \) OL\(^{2}\) = KL\(^{2}\) + OK\(^{2}\) [Pythagoras theorem]
\( \therefore \) 5\(^{2}\) = 3.5\(^{2}\) + OK\(^{2}\)
\( \therefore \) OK\(^{2}\) = 25 - 12.25 = 12.75 Now, in \(\triangle\)OKM, \(\angle\)OKM = 90°
\( \therefore \) OM\(^{2}\) = OK\(^{2}\) + MK\(^{2}\) = 12.75 + 12.5\(^{2}\) = 12.75 + 156.25 = 169
\( \therefore \) OM = \( \sqrt{169} \) = 13 units [Taking square root of both sides]
\( \therefore \) The radius of the given circle is 13 units. In simple words: To find the radius, we construct a perpendicular from the center to the chord, which bisects the chord. Then, by applying the Pythagorean theorem in the formed right-angled triangles using the given distances and calculated segments, we can determine the radius.

🎯 Exam Tip: Break down complex geometry problems into smaller right-angled triangles. The Pythagorean theorem is a powerful tool. Ensure calculations for squares and square roots are accurate.

Question 22. In the adjoining figure, two circles intersect each other at points S and R. Their common tangent PQ touches the circle at points P, Q. Prove that, \(\angle\)PRQ + \(\angle\)PSQ = 180°.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो वृत्तों को दर्शाता है जो बिंदु S और R पर एक-दूसरे को प्रतिच्छेद करते हैं। एक उभयनिष्ठ स्पर्शरेखा PQ है जो वृत्तों को क्रमशः P और Q पर स्पर्श करती है। यह आकृति स्पर्शरेखा-जीवा प्रमेय और त्रिभुज के कोण योग गुण का उपयोग करके कोणों के संबंधों को सिद्ध करती है।
Answer: Given: Two circles intersect each other at points S and R. line PQ is a common tangent. To prove: \(\angle\)PRQ + \(\angle\)PSQ = 180° Proof: Line PQ is the tangent at point P and seg PR is a secant.
\( \therefore \) \(\angle\)RPQ = \(\angle\)PSR ............. (i) and \(\angle\)PQR = \(\angle\)QSR] ............ (ii) [Tangent secant theorem] In \(\triangle\)PQR, \(\angle\)PQR + \(\angle\)PRQ + \(\angle\)RPQ = 180° [Sum of the measures of angles of a triangle is 180°]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो प्रतिच्छेद करने वाले वृत्तों को दिखाता है, जिनके प्रतिच्छेदन बिंदु S और R हैं। रेखा PQ एक उभयनिष्ठ स्पर्शरेखा है जो P और Q पर वृत्तों को स्पर्श करती है। विभिन्न जीवाएँ खींची गई हैं जो कोणों \(\angle\)RPQ, \(\angle\)PSR, \(\angle\)PQR, \(\angle\)QSR को परिभाषित करती हैं।
Answer:
\( \therefore \) \(\angle\)QSR + \(\angle\)PRQ + \(\angle\)PSR = 180° [From (i) and (ii)]
\( \therefore \) \(\angle\)PRQ + \(\angle\)QSR + \(\angle\)PSR = 180°
\( \therefore \) \(\angle\)PRQ + \(\angle\)PSQ = 180° [Angle addition property] In simple words: By applying the tangent-secant theorem to both circles, we can replace angles in the triangle sum with angles related to the second circle. Then, using angle addition, we prove that the sum of the required angles is 180 degrees.

🎯 Exam Tip: The tangent-secant theorem is frequently tested. Clearly state which angle is equal to which angle in the alternate segment. Breaking down the proof into steps for each circle helps manage complexity.

Question 23. In the adjoining figure, two circles intersect at points M and N. Secants drawn through M and N intersect the circles at points R, S and P, Q respectively. Prove that: seg SQ || seg RP.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो वृत्तों को दर्शाता है जो बिंदु M और N पर प्रतिच्छेद करते हैं। दो छेदक रेखाएँ खींची गई हैं। पहली छेदक रेखा M से गुजरकर वृत्त को R और P पर काटती है। दूसरी छेदक रेखा N से गुजरकर वृत्त को S और Q पर काटती है।
Answer: Given: Two circles intersect each other at points M and N. To prove: seg SQ || seg RP Construction: Join seg MN. Proof: RMNP is a cyclic quadrilateral.
\( \therefore \) \(\angle\)MRP = \(\angle\)MNQ ............ (i) [Corollary of cyclic quadrilateral theorem] Also, MNQS is a cyclic quadrilateral.
\( \therefore \) \(\angle\)MNQ+ \(\angle\)MSQ = 180° [Theorem of cyclic quadrilateral]
\( \therefore \) \(\angle\)MRP + \(\angle\)MSQ = 180° [From (i)]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो प्रतिच्छेद करने वाले वृत्तों को दर्शाता है। बिंदु M और N प्रतिच्छेदन बिंदु हैं। छेदक रेखाएँ RMP और SNQ हैं। जीवा MN को जोड़ा गया है। यह आकृति दिखाती है कि कैसे \(\angle\)MRP और \(\angle\)MSQ के बीच संबंध स्थापित करके रेखाओं SQ और RP को समांतर सिद्ध किया जा सकता है।
Answer: But, they are a pair of interior angles on the same side of transversal RS on lines SQ and RP.
\( \therefore \) seg SQ || seg RP [Interior angles test] In simple words: By constructing a common chord (MN), we form two cyclic quadrilaterals. Using the property that the exterior angle of a cyclic quadrilateral is equal to its interior opposite angle, and opposite angles sum to 180 degrees, we can establish angle relationships that prove the lines SQ and RP are parallel by the interior angles test.

🎯 Exam Tip: Auxiliary constructions (like joining MN here) are often key in proofs. Remember the properties of cyclic quadrilaterals, especially regarding exterior angles and opposite angles.

Question 24. In the adjoining figure, two circles intersect each other at points A and E. Their common secant through E intersects the circles at points B and D. The tangents of the circles at points B and D intersect each other at point C. Prove that \(\Box\)ABCD is cyclic.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो प्रतिच्छेद करने वाले वृत्तों को दर्शाता है, जिनके प्रतिच्छेदन बिंदु A और E हैं। एक उभयनिष्ठ छेदक रेखा B-E-D दोनों वृत्तों को काटती है। बिंदु B और D पर खींची गई स्पर्शरेखाएँ बिंदु C पर प्रतिच्छेद करती हैं। यह आकृति चतुर्भुज ABCD को चक्रीय सिद्ध करने के लिए कोणों के संबंधों को दर्शाती है।
Answer: Given: Two circles intersect each other at A and E. seg BC and seg CD are the tangents to the circles. To prove: \(\Box\)ABCD is cyclic. Construction: Draw AB, AE and AD. Proof: \(\angle\)EBC = \(\angle\)BAE (i)
\(\angle\)EDC = \(\angle\)DAE ] (ii) [Tangent secant theorem] In \(\triangle\)BCD, \(\angle\)DBC + \(\angle\)BDC + \(\angle\)BCD = 180° [Sum of the measures of angles of a triangle is 180°]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो वृत्तों को दर्शाता है जो बिंदु A और E पर प्रतिच्छेद करते हैं। एक छेदक रेखा BED है। बिंदु B और D पर खींची गई स्पर्शरेखाएँ C पर मिलती हैं, जिससे चतुर्भुज ABCD बनता है। यह आकृति स्पर्शरेखा-जीवा प्रमेय का उपयोग करके \(\angle\)EBC, \(\angle\)BAE, \(\angle\)EDC और \(\angle\)DAE के बीच संबंध स्थापित करती है।
Answer:
\( \therefore \) \(\angle\)EBC + \(\angle\)EDC + \(\angle\)BCD = 180° (iii) [B - E - D]
\( \therefore \) \(\angle\)BAE + \(\angle\)DAE + \(\angle\)BCD = 180° [From (i), (ii) and (iii)]
\( \therefore \) \(\angle\)BAD + \(\angle\)BCD = 180° [Angle addition property]
\( \therefore \) \(\Box\)ABCD is cyclic. [Converse of cyclic quadrilateral theorem] In simple words: By using the tangent-secant theorem and the angle sum property of a triangle, we substitute angles to show that the sum of opposite angles of quadrilateral ABCD (\(\angle\)BAD and \(\angle\)BCD) is 180 degrees. This directly proves that ABCD is a cyclic quadrilateral by the converse of the cyclic quadrilateral theorem.

🎯 Exam Tip: The converse of the cyclic quadrilateral theorem is a powerful tool for proving cyclicity. Ensure all substitutions and angle additions are clearly justified and follow logically from the theorems applied.

Question 25. In the adjoining figure, seg AD \( \perp \) side BC, seg BE \( \perp \) side AC, seg CF \( \perp \) side AB. Point O is the orthocentre. Prove that, point O is the incentre of \(\triangle\)DEF.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ABC को दर्शाता है जिसमें AD, BE, CF शीर्षलंब हैं, जो बिंदु O (लंबकेंद्र) पर प्रतिच्छेद करते हैं। बिंदु D, E, F भुजाओं BC, AC, AB पर स्थित हैं। यह आकृति सिद्ध करती है कि लंबकेंद्र O, शीर्षलंबों के पाद-बिंदुओं से बने त्रिभुज DEF का अंतःकेंद्र होता है।
Answer: Given: seg AD \( \perp \) side BC, seg BE \( \perp \) side AC, seg CF \( \perp \) side AB. Point O is the orthocentre. To prove: Point O is the incentre of \(\triangle\)DEF. Construction: Draw DE, EF and DF. Proof: \(\angle\)OFA = \(\angle\)OEA = 90° [Given]
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक त्रिभुज ABC है जिसके शीर्षलंब AD, BE, CF हैं, जो बिंदु O पर प्रतिच्छेद करते हैं। बिंदु D, E, F भुजाओं BC, AC, AB पर स्थित हैं। इस आकृति में कई चक्रीय चतुर्भुज (जैसे OFAE, OFBD) बनते हैं, जिनका उपयोग यह सिद्ध करने के लिए किया जाता है कि O, त्रिभुज DEF का अंतःकेंद्र है।
Answer: Now, \(\angle\)OFA + \(\angle\)OEA = 90° + 90°
\( \therefore \) \(\angle\)OFA + \(\angle\)OEA = 180°
\( \therefore \) \(\Box\)OFAE is a cyclic quadrilateral. [Converse of cyclic quadrilateral theorem]
\( \therefore \) Points O, F, A, E are concyclic points.
\( \therefore \) seg OE subtends equal angles \(\angle\)OFE and \(\angle\)OAE on the same side of OЕ.
\( \therefore \) \(\angle\)OFE = \(\angle\)OAE ............. (i) \(\angle\)OFB = \(\angle\)ODB = 90° [Given] Now, \(\angle\)OFB + \(\angle\)ODB = 90° + 90°
\( \therefore \) \(\angle\)OFB + \(\angle\)ODB = 180°
\( \therefore \) \(\Box\)OFBD is a cyclic quadrilateral. [Converse of cyclic quadrilateral theorem]
\( \therefore \) Points O, F, B, D are concyclic points.
\( \therefore \) seg OD subtends equal angles \(\angle\)OFD and \(\angle\)OBD on the same side of OD. \(\angle\)OFD = \(\angle\)OBD ............. (ii) In \(\triangle\)AEO and \(\triangle\)BDO, \(\angle\)AEO = \(\angle\)BDO [Each angle is 90°] \(\angle\)AOE = \(\angle\)BOD [Vertically opposite angles]
\( \therefore \) \(\triangle\)AEO \( \sim \) \(\triangle\)BDO [AA test of similarity]
\( \therefore \) \(\angle\)OAE = \(\angle\)OBD ............ (iii) [Corresponding angles of similar triangles]
\( \therefore \) \(\angle\)OFE = \(\angle\)OFD [From (i), (ii) and (iii)]
\( \therefore \) ray FO bisects \(\angle\)EFD. Similarly, we can prove ray EO and ray DO bisects \(\angle\)FED and \(\angle\)FDE respectively.
\( \therefore \) Point O is the intersection of angle bisectors of \(\angle\)D, \(\angle\)E and \(\angle\)F of \(\triangle\)DEF.
\( \therefore \) Point O is the incentre of \(\triangle\)DEF. In simple words: By identifying cyclic quadrilaterals formed by the orthocenter and the feet of altitudes, we show that angles subtended by the same arc are equal. Using similarity and angle relationships, we prove that the orthocenter bisects the angles of the pedal triangle (formed by points D, E, F), thus making it the incenter of \(\triangle\)DEF.

🎯 Exam Tip: This is a complex proof involving multiple geometric properties. Systematically proving cyclic quadrilaterals and using their angle properties, along with similarity, will lead to the solution. Clearly state all theorems used.