Maharashtra Board Class 10 Maths Chapter 3 Circle Set 3.5 Solutions

Question 1. In the adjoining figure, ray PQ touches the circle at point Q. PQ = 12, PR = 8, find PS and RS.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त को स्पर्श करती हुई एक किरण PQ है, जो बिंदु Q पर स्पर्श करती है। एक छेदक रेखा PRS है जो वृत्त को R और S बिंदुओं पर काटती है, और बिंदु P वृत्त के बाहर स्थित है।
Answer:
Solution:
i. Ray PQ is a tangent to the circle at point Q and seg PS is the secant. [Given]
\( \therefore \) PR \( \times \) PS = PQ\(^2\) [Tangent secant segments theorem]
\( \therefore \) 8 \( \times \) PS = 12\(^2\)
\( \therefore \) 8 \( \times \) PS = 144
\( \therefore \) PS = \(\frac{144}{8}\)
\( \therefore \) PS = 18 units
ii. Now, PS = PR + RS [P – R – S]
\( \therefore \) 18 = 8 + RS
\( \therefore \) RS = 18 - 8
\( \therefore \) RS = 10 units
In simple words: Using the tangent-secant theorem, we first found the length of PS. Then, by using the collinearity of points P, R, and S, we calculated the length of RS by subtracting PR from PS.

🎯 Exam Tip: Remember the tangent-secant theorem \(PQ^2 = PR \times PS\) for external point P, tangent PQ, and secant PRS. Also, pay attention to the collinearity of points for segment addition/subtraction.

Question 2. In the adjoining figure, chord MN and chord RS intersect at point D.
i. If RD = 15, DS = 4, MD = 8 find DN
ii. If RS = 18, MD = 9, DN = 8 find DS
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त के अंदर दो जीवाएँ MN और RS बिंदु D पर एक-दूसरे को काटती हैं। बिंदु M, N, R, S वृत्त की परिधि पर स्थित हैं।
Answer:
Solution:
i. Chords MN and RS intersect internally at point D. [Given]
\( \therefore \) MD \( \times \) DN = RD \( \times \) DS [Theorem of internal division of chords]
\( \therefore \) 8 \( \times \) DN = 15 \( \times \) 4
\( \therefore \) DN = \(\frac{15 \times 4}{8}\)
\( \therefore \) DN = 7.5 units
ii. Let the value of RD be x.
RS = RD + DS [R – D – S]
\( \therefore \) 18 = x + DS
\( \therefore \) DS = 18 - x
Now, MD \( \times \) DN = RD \( \times \) DS [Theorem of internal division of chords]
\( \therefore \) 9 \( \times \) 8 = x(18 - x)
\( \therefore \) 72 = 18x - x\(^2\)
\( \therefore \) x\(^2\) - 18x + 72 = 0
\( \therefore \) x\(^2\) - 12x - 6x + 72 = 0
\( \therefore \) x (x - 12) - 6 (x - 12) = 0
\( \therefore \) (x - 12) (x - 6) = 0
\( \therefore \) x - 12 = 0 or x - 6 = 0
\( \therefore \) x = 12 or x = 6
\( \therefore \) DS = 18 - 12 or DS = 18 - 6
\( \therefore \) DS = 6 units or DS = 12 units
In simple words: This problem uses the theorem of internal division of chords. For part (i), we directly applied \(MD \times DN = RD \times DS\) to find DN. For part (ii), we used the same theorem after setting up an equation for the lengths based on RS and solving the quadratic equation for x (RD) and then finding DS.

🎯 Exam Tip: The theorem of internal division of chords states that for chords AB and CD intersecting at D, \(AD \times DB = CD \times DE\). Be careful with quadratic equations and potential multiple solutions for lengths, ensuring they are valid in context.

Question 3. In the adjoining figure, O is the centre of the circle and B is a point of contact. Seg OE \( \perp \) seg AD, AB = 12, AC = 8, find
i. AD
ii. DC
iii. DE.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त है जिसका केंद्र O है। बिंदु B पर एक स्पर्श रेखा AB वृत्त को छूती है। एक छेदक रेखा ACD वृत्त को C और D बिंदुओं पर काटती है, जहाँ A वृत्त के बाहर है। रेखा खंड OE, रेखा खंड AD पर लंबवत है, जहाँ E जीवा CD पर स्थित है।
Answer:
Solution:
i. Line AB is the tangent at point B and seg AD is the secant. [Given]
\( \therefore \) AC \( \times \) AD = AB\(^2\) [Tangent secant segments theorem]
\( \therefore \) 8 \( \times \) AD = 12\(^2\)
\( \therefore \) 8 \( \times \) AD = 144
\( \therefore \) AD = \(\frac{144}{8}\)
\( \therefore \) AD = 18 units
ii. AD = AC + DC [A – C – D]
\( \therefore \) 18 = 8 + DC
\( \therefore \) DC = 18 - 8
\( \therefore \) DC = 10 units
iii. seg OE \( \perp \) seg AD [Given]
i.e. seg OE \( \perp \) seg CD [A – C – D]
\( \therefore \) DE = \(\frac{1}{2}\) DC [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
= \(\frac{1}{2}\) \( \times \) 10
\( \therefore \) DE = 5 units
In simple words: We used the tangent-secant theorem to find AD. Then, using segment addition, we found DC. Finally, since a perpendicular from the center bisects a chord, we found DE as half of DC.

🎯 Exam Tip: This problem combines the tangent-secant theorem with the property that a perpendicular from the center to a chord bisects the chord. Ensure you correctly apply each geometric theorem in the right step.

Question 4. In the adjoining figure, if PQ = 6, QR = 10, PS = 8, find TS.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त के बाहर बिंदु P से दो छेदक रेखाएँ खींची गई हैं। पहली छेदक रेखा P-Q-R है जो वृत्त को Q और R पर काटती है। दूसरी छेदक रेखा P-S-T है जो वृत्त को S और T पर काटती है।
Answer:
Solution:
PR = PQ + QR [P-Q-R]
\( \therefore \) PR = 6 + 10 = 16 units
Chords TS and RQ intersect externally at point P.
PQ \( \times \) PR = PS \( \times \) PT [Theorem of external division of chords]
\( \therefore \) 6 \( \times \) 16 = 8 \( \times \) PT
\( \therefore \) PT = \(\frac{6 \times 16}{8}\) = 12 units
But, PT = PS + TS [P – S – T]
\( \therefore \) 12 = 8 + TS
\( \therefore \) TS = 12 - 8
\( \therefore \) TS = 4 units
In simple words: We first found the total length of the secant segment PR. Then, we applied the theorem of external division of chords (\(PQ \times PR = PS \times PT\)) to find PT. Finally, we used segment subtraction to find TS.

🎯 Exam Tip: The theorem of external division of chords is crucial here. Remember to correctly identify the full length of the secant segments (e.g., PR and PT) and the external portions (PQ and PS).

Question 5. In the adjoining figure, seg EF is a diameter and seg DF is a tangent segment. The radius of the circle is r. Prove that, DE \( \times \) GE = 4r\(^2\).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त है जिसका व्यास EF है। बिंदु D वृत्त के बाहर स्थित है और DF वृत्त पर F बिंदु पर एक स्पर्श रेखा है। G वृत्त पर एक बिंदु है। रेखाखंड DE, G से होकर गुजरता है।
Answer:
Given: seg EF is the diameter.
seg DF is a tangent to the circle,
radius = r
To prove: DE \( \times \) GE = 4r\(^2\)
Construction: Join seg GF.
Proof:
seg EF is the diameter. [Given]
\( \therefore \) \( \angle \)EGF = 90\(^\circ\) (i) [Angle inscribed in a semicircle]
seg DF is a tangent to the circle at F. [Given]
\( \therefore \) \( \angle \)EFD = 90\(^\circ\) (ii) [Tangent theorem]
In \( \triangle \)DFE,
\( \angle \)EFD = 90 \(^\circ\) [From (ii)]
seg FG \( \perp \) side DE [From (i)]
\( \therefore \) \( \triangle \)EFD \( \sim \) \( \triangle \)EGF [Similarity of right angled triangles]
\( \therefore \frac{DE}{EF} = \frac{EF}{GE} \) [Corresponding sides of similar triangles]
\( \therefore \) DE \( \times \) GE = EF\(^2\)
\( \therefore \) DE \( \times \) GE = (2r)\(^2\) [diameter = 2r]
\( \therefore \) DE \( \times \) GE = 4r\(^2\)
In simple words: We first established that \( \angle \)EGF and \( \angle \)EFD are 90 degrees due to the angle in a semicircle and the tangent theorem, respectively. This allowed us to prove similarity between \( \triangle \)EFD and \( \triangle \)EGF, leading to the proportion \(DE/EF = EF/GE\). From this, we derived \(DE \times GE = EF^2\), and since EF is the diameter (2r), the final result is \(DE \times GE = 4r^2\).

🎯 Exam Tip: This proof requires the application of multiple theorems: angle in a semicircle, tangent theorem, and similarity of right-angled triangles. Clearly state the reasons for each step to score full marks.

Question 1. Theorem: If an angle has its vertex on the circle, its one side touches the circle and the other intersects the circle in one more point, then the measure of the angle is half the measure of its intercepted arc. (Textbook pg.no. 75 and 76)
ℹ️ चित्र व्याख्या (Diagram Explanation):
(i) एक वृत्त जिसका केंद्र M है। एक स्पर्श रेखा BC बिंदु B पर स्पर्श करती है। एक छेदक रेखा BA वृत्त को बिंदु A पर काटती है। कोण ABC दर्शाया गया है।
(ii) एक वृत्त जिसका केंद्र M है। एक स्पर्श रेखा BC बिंदु B पर स्पर्श करती है। एक छेदक रेखा BA वृत्त को बिंदु A पर काटती है। केंद्र M से A और B तक त्रिज्याएँ खींची गई हैं, जिससे त्रिभुज MAB बनता है।
(iii) एक वृत्त जिसका केंद्र M है। बिंदु F वृत्त की परिधि पर है, और एक छेदक रेखा EFA है। कोण ABE दर्शाया गया है। वृत्त का केंद्र M कोण ABC के आंतरिक भाग में स्थित है।
Answer:
Given: \( \angle \)ABC is any angle, whose vertex B lies on the circle with centre M.
Line BC is tangent at B and line BA is secant intersecting the circle at point A.
Arc ADB is intercepted by \( \angle \)ABC.
To prove: \( \angle \)ABC = \(\frac{1}{2}\) m(arc ADB)
Proof:
Case I: Centre M lies on arm BA of \( \angle \)ABC.
\( \angle \)MBC = 90\(^\circ\) [Tangent theorem]
i.e. \( \angle \)ABC = 90\(^\circ\) (i) [A – M – B]
arc ADB is a semicircular arc.
\( \therefore \) m(arc ADB) = 180\(^\circ\) (ii) [Measure of a semicircle is 180\(^\circ\)]
\( \therefore \) \( \angle \)ABC = \(\frac{1}{2}\) m(arc ADB) [(From (i) and (ii)]
Case II: Centre M lies in the exterior of \( \angle \)ABC.
Draw radii MA and MB.
\( \therefore \) \( \angle \)MBA = \( \angle \)MAB [Isosceles triangle theorem]
Let, \( \angle \)MHA = \( \angle \)MAB =x, \( \angle \)ABC = y In \( \triangle \)ABM,
\( \angle \)AMB + \( \angle \)MBA + \( \angle \)MAB = 180\(^\circ\) [Sum of the measures of all the angles of a triangle is 180\(^\circ\)]
\( \therefore \) \( \angle \)AMB + x + x = 180\(^\circ\)
\( \therefore \) \( \angle \)AMB = 180\(^\circ\) - 2x ...... (i)
Now, \( \angle \)MBC = \( \angle \)MBA + \( \angle \)ABC [Angle addition property]
\( \therefore \) 90\(^\circ\) = x + y [Tangent theorem]
\( \therefore \) x = 90\(^\circ\) - y ......(ii)
Substituting (ii) in (i),
\( \angle \)AMB = 180\(^\circ\) - 2 (90\(^\circ\) - y) [From (i) and (ii)]
\( \therefore \) \( \angle \)AMB = 180\(^\circ\) - 180\(^\circ\) + 2y
\( \therefore \) 2y = \( \angle \)AMB
\( \therefore \) y = \(\frac{1}{2}\) \( \angle \)AMB
\( \therefore \) \( \angle \)ABC = \(\frac{1}{2}\) \( \angle \)AMB
\( \therefore \) \( \angle \)ABC = \(\frac{1}{2}\) m(arc ADB) [Definition of measure of minor arc]
Case III: Centre M lies in the interior of \( \angle \)ABC.
Ray BE is the opposite ray of ray BC.
Now, \( \angle \)ABE = \(\frac{1}{2}\) m (arc AFB) (i) [Proved in case II]
\( \angle \)ABC + \( \angle \)ABE = 180\(^\circ\) [Angles in a linear pair]
\( \therefore \) 180 - \( \angle \)ABC = \( \angle \)ABE
\( \therefore \) 180 - \( \angle \)ABC = \(\frac{1}{2}\) m(arc AFB) [From (i)]
= \(\frac{1}{2}\) [360 - m (arc ADB)]
\( \therefore \) 180 - \( \angle \)ABC = 180 - \(\frac{1}{2}\) m(arc ADB)
\( \therefore \) -\( \angle \)ABC = -\(\frac{1}{2}\) m(arc ADB)
\( \therefore \) \( \angle \)ABC = \(\frac{1}{2}\) m(arc ADB)
In simple words: This theorem proves that the angle formed by a tangent and a secant at the point of contact is half the measure of its intercepted arc. The proof is broken into three cases depending on the position of the circle's center relative to the angle, using tangent theorems, angle sum property of triangles, and definitions of arc measures.

🎯 Exam Tip: This is a fundamental theorem. Understand the three cases (center on arm, exterior, interior) and how tangent properties, isosceles triangles, and angle relationships are used to prove each case. Drawing accurate diagrams for each case helps in visualization.

Question 2. We have proved the above theorem by drawing seg AC and seg DB. Can the theorem be proved by drawing seg AD and seg CB, instead of seg AC and seg DB? (Textbook pg. no. 77)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त के अंदर दो जीवाएँ AB और CD बिंदु E पर एक-दूसरे को काटती हैं। बिंदु A, B, C, D वृत्त की परिधि पर स्थित हैं। बिंदु P वृत्त का केंद्र है।
Answer:
Solution:
Yes, the theorem can be proved by drawing seg AD and seg CB.
Given: P is the centre of circle, chords AB and CD intersect internally at point E.
To prove: AE \( \times \) EB = CE \( \times \) ED
Construction: Draw seg AD and seg CB.
Proof:
In \( \triangle \)CEB and \( \triangle \)AED,
\( \angle \)CEB = \( \angle \)DEA [Vertically opposite angles]
\( \angle \)CBE = \( \angle \)ADE [Angles inscribed in the same arc]
\( \therefore \) \( \triangle \)CEB \( \sim \) \( \triangle \)AED [by AA test of similarity]
\( \therefore \frac{CE}{AE} = \frac{ED}{EB} \) [Corresponding sides of similar triangles]
\( \therefore \) AE \( \times \) EB = CE \( \times \) ED
In simple words: Yes, the theorem of internal division of chords can be proved by drawing AD and CB. By showing that \( \triangle \)CEB is similar to \( \triangle \)AED using vertically opposite angles and angles in the same arc, the proportionality of their sides directly leads to the desired product relation.

🎯 Exam Tip: The key to proving this theorem is establishing similarity between the two triangles formed by the intersecting chords. Vertically opposite angles and angles subtended by the same arc are essential criteria for AA similarity.

Question 3. In figure, seg PQ is a diameter of a circle with centre O. R is any point on the circle, seg RS \( \perp \) seg PQ. Prove that, SR is the geometric mean of PS and SQ. [That is, SR\(^2\) = PS \( \times \) SQ] (Textbook pg. no. 81)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त है जिसका केंद्र O और व्यास PQ है। वृत्त की परिधि पर एक बिंदु R है। एक रेखाखंड RS, PQ पर S बिंदु पर लंबवत है। किरण RS को आगे बढ़ाया गया है ताकि यह वृत्त को T बिंदु पर काटे।
Answer:
Given: seg PQ is the diameter.
seg RS \( \perp \) seg PQ
To prove: SR\(^2\) = PS \( \times \) SQ
Construction: Extend ray RS, let it intersect the circle at point T.
Proof:
seg PQ \( \perp \) seg RS [Given]
\( \therefore \) seg OS \( \perp \) chord RT [R – S – T, P – S – O]
\( \therefore \) segSR = segTS (i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
Chords PQ and RT intersect internally at point S.
\( \therefore \) SR \( \times \) TS = PS \( \times \) SQ [Theorem of internal division of chords]
\( \therefore \) SR \( \times \) SR = PS \( \times \) SQ [From (i)]
\( \therefore \) SR\(^2\) = PS \( \times \) SQ
In simple words: To prove SR is the geometric mean, we extended RS to T and used the property that a perpendicular from the center bisects a chord, making SR = TS. Then, applying the theorem of internal division of chords to chords PQ and RT intersecting at S, we substituted SR for TS to derive \(SR^2 = PS \times SQ\).

🎯 Exam Tip: This problem beautifully combines two key theorems: the property of a perpendicular from the center to a chord and the theorem of internal division of chords. The construction step of extending RS to T is critical.

Question 4. Theorem: If secants containing chords AB and CD of a circle intersect outside the circle in point E, then AE \( \times \) EB = CE \( \times \) ED. (Textbook pg. no. 78)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त है जिसके बाहर बिंदु E से दो छेदक रेखाएँ खींची गई हैं। एक छेदक रेखा E-A-B वृत्त को A और B पर काटती है। दूसरी छेदक रेखा E-C-D वृत्त को C और D पर काटती है।
Answer:
Given: Chords AB and CD of a circle intersect outside the circle in point E.
To prove: AE \( \times \) EB = CE \( \times \) ED
Construction: Draw seg AD and seg BC.
Proof:
In \( \triangle \)ADE and \( \triangle \)CBE,
\( \angle \)AED = \( \angle \)CEB [Common angle]
\( \angle \)DAE = \( \angle \)BCE [Angles inscribed in the same arc]
\( \therefore \) \( \triangle \)ADE \( \sim \) \( \triangle \)CBE [AA test of similarity]
\( \therefore \frac{AE}{CE} = \frac{ED}{EB} \) [Corresponding sides of similar triangles]
\( \therefore \) AE \( \times \) EB = CE \( \times \) ED
In simple words: This theorem, known as the theorem of external division of chords, is proved by demonstrating the similarity between \( \triangle \)ADE and \( \triangle \)CBE. The common angle E and inscribed angles subtended by the same arc (DA and CB) are used to establish AA similarity, leading to the proportionality of sides and thus the product relation.

🎯 Exam Tip: This is a standard theorem of intersecting secants outside a circle. The key is to draw auxiliary lines (AD and BC) to form similar triangles and then use the property of corresponding sides to establish the product relationship.

Question 5. Theorem: Point E is in the exterior of a circle. A secant through E intersects the circle at points A and B, and a tangent through E touches the circle at point T, then EA \( \times \) EB = ET\(^2\).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त है जिसके बाहर बिंदु E है। एक छेदक रेखा EAB वृत्त को A और B बिंदुओं पर काटती है। बिंदु E से खींची गई एक स्पर्श रेखा ET वृत्त को T बिंदु पर स्पर्श करती है।
Answer:
Given: Secant through point E intersects the circle in points A and B.
Tangent drawn through point E touches the circle in point T.
To prove: EA \( \times \) EB = ET\(^2\)
Construction: Draw seg TA and seg TB.
Proof:
In \( \triangle \)EAT and \( \triangle \)ETB,
\( \angle \)AET = \( \angle \)TEB [Common angle]
\( \angle \)ETA = \( \angle \)EBT [Theorem of angle between tangent and secant, E – A – B]
\( \therefore \) \( \triangle \)EAT \( \sim \) \( \triangle \)ETB [AA test of similarity]
\( \therefore \frac{EA}{ET} = \frac{ET}{EB} \) [Corresponding sides of similar triangles]
\( \therefore \) EA \( \times \) EB = ET\(^2\)
In simple words: This theorem, also known as the tangent-secant theorem, proves that the square of the length of the tangent segment from an external point is equal to the product of the lengths of the secant segment and its external part. It is proved by showing the similarity between \( \triangle \)EAT and \( \triangle \)ETB using the common angle E and the angle between tangent and secant.

🎯 Exam Tip: This is the fundamental tangent-secant theorem. The construction of TA and TB is crucial to form the similar triangles. Remember that the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

Question 6. In the figure in the above example, if seg PR and seg RQ are drawn, what is the nature of \( \triangle \)PRQ. (Textbook pg. no, 81)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त है जिसका व्यास PQ है। वृत्त की परिधि पर एक बिंदु R है। एक रेखाखंड RS, PQ पर S बिंदु पर लंबवत है। किरण RS को आगे बढ़ाया गया है ताकि यह वृत्त को T बिंदु पर काटे। त्रिभुज PRQ को दर्शाया गया है।
Answer:
seg PQ is the diameter of the circle.
\( \therefore \) \( \angle \)PRQ = 90\(^\circ\)
\( \therefore \) \( \triangle \)PRQ is a right angled triangle. [Angle inscribed in a semicircle]
In simple words: Since PQ is the diameter of the circle and R is a point on the circle, the angle \( \angle \)PRQ is an angle inscribed in a semicircle. An angle inscribed in a semicircle is always a right angle, thus \( \triangle \)PRQ is a right-angled triangle.

🎯 Exam Tip: This is a direct application of the property that an angle inscribed in a semicircle is always a right angle (90\(^\circ\)). This is a frequently tested concept.

Question 7. Have you previously proved the property proved in the above example? (Textbook pg. no. 81)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त है जिसका व्यास PQ है। वृत्त की परिधि पर एक बिंदु R है। एक रेखाखंड RS, PQ पर S बिंदु पर लंबवत है। किरण RS को आगे बढ़ाया गया है ताकि यह वृत्त को T बिंदु पर काटे।
Answer:
Yes. It is the theorem of geometric mean.
\( \triangle \)PSR \( \sim \) \( \triangle \)RSQ [Similarity of right angled triangles]
\( \therefore \frac{PS}{SR} = \frac{SR}{SQ} \) [Corresponding sides of similar triangles]
\( \therefore \) SR\(^2\) = PS \( \times \) SQ
In simple words: Yes, this property is the theorem of geometric mean, which states that in a right-angled triangle, the altitude to the hypotenuse is the geometric mean of the two segments it divides the hypotenuse into. This is proved by showing the similarity between \( \triangle \)PSR and \( \triangle \)RSQ.

🎯 Exam Tip: The geometric mean theorem (also known as the altitude rule) is a special case of similar triangles in a right-angled triangle when an altitude is drawn to the hypotenuse. Recognizing this connection is key.