Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 6 Set 6.2 Circle Solutions

Question 1. Radius of circle is 10 cm. There are two chords of length 16 cm each. What will be the distance of these chords from the centre of the circle ?
Given: In a circle with centre O,
OR and OP are radii and RS and PQ are its congruent chords.
PQ = RS= 16 cm,
OR = OP = 10 cm
seg OU \( \perp \) chord PQ, P-U-Q
seg OT \( \perp \) chord RS, R-T-S
To find: Distance of chords from centre of the circle.
Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त का चित्र है जिसका केंद्र O है। इसमें दो जीवाएँ PQ और RS दिखाई गई हैं। केंद्र O से PQ पर OU लंब है और RS पर OT लंब है, जो जीवाओं को समद्विभाजित करते हैं। वृत्त की त्रिज्याएँ OP और OR 10 सेमी हैं, और दोनों जीवाओं की लंबाई 16 सेमी है।
i. PU = \( \frac{1}{2} \)(PQ) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
\( \therefore \) PU= \( \frac{1}{2} \) x 16 = 8 cm ...(i)

ii. In \( \Delta \)OUP, \( \angle \)OUP = 90°
\( \therefore \) OP\(^2\) = OU\(^2\) + PU\(^2\) [Pythagoras theorem]
\( \therefore \) 10\(^2\) = OU\(^2\) + 8\(^2\) [From (i)]
\( \therefore \) 100 = OU\(^2\) + 64
\( \therefore \) OU\(^2\) = 100 - 64 = 36
\( \therefore \) OU = \( \sqrt{36} \) [Taking square root on both sides]
\( \therefore \) OU = 6 cm

iii. Now, OT = OU [Congruent chords of a circle are equidistant from the centre.]
\( \therefore \) OT = OU = 6cm
\( \therefore \) The distance of the chords from the centre of the circle is 6 cm.
Answer: The distance of the chords from the centre of the circle is 6 cm.
In simple words: Since congruent chords are equidistant from the center, and the radius and half-chord form a right-angled triangle, we use the Pythagorean theorem. With a radius of 10 cm and a half-chord of 8 cm, the distance from the center is calculated as 6 cm.

🎯 Exam Tip: Remember that a perpendicular from the center of a circle to a chord bisects the chord. This, along with the Pythagorean theorem, is crucial for solving problems involving chord lengths and distances from the center.

Question 2. In a circle with radius 13 cm, two equal chords are at a distance of 5 cm from the centre. Find the lengths of chords.
Given: In a circle with cente O,
OA and OC are the radii and AB and CD are its congruent chords,
OA = OC = 13cm
OE = OF = 5 cm
seg OE \( \perp \) chord CD, C-E-D
seg OF \( \perp \) chord AB. A-F-B
To find: length of the chords
Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त का चित्र है जिसका केंद्र O है। इसमें दो समान जीवाएँ AB और CD दिखाई गई हैं। केंद्र O से जीवा CD पर OE लंब है और जीवा AB पर OF लंब है। वृत्त की त्रिज्याएँ OA और OC 13 सेमी हैं, और केंद्र से जीवाओं की दूरी OE और OF 5 सेमी है।
i. In \( \Delta \)AFO, \( \angle \)AFO = 90°
\( \therefore \) AO\(^2\) = AF\(^2\) + FO\(^2\) [Pythagoras theorem]
\( \therefore \) 13\(^2\) = AF\(^2\) + 5\(^2\)
\( \therefore \) 169 = AF\(^2\) + 25
\( \therefore \) AF\(^2\) = 169-25
\( \therefore \) AF\(^2\) = 144
\( \therefore \) AF = \( \sqrt{144} \) [Taking square root on both sides]
\( \therefore \) AF = 12 cm .....(i)

ii. Now AF = \( \frac{1}{2} \) AB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
\( \therefore \) 12 = \( \frac{1}{2} \)(AB) [From (i)]
\( \therefore \) AB = 12 x 2 = 24 cm
\( \therefore \) CD = AB = 24 cm [chord AB \( \simeq \) chord CD]
\( \therefore \) The lengths of the two chords are 24 cm each.
Answer: The lengths of the two chords are 24 cm each.
In simple words: Given the radius and the distance of equal chords from the center, we use the Pythagorean theorem to find half the length of one chord. Since the chords are equal, their lengths are identical. Double the half-chord length gives the full chord length.

🎯 Exam Tip: Always remember that chords equidistant from the center of a circle are congruent. This property, combined with the perpendicular bisector theorem and Pythagoras, is fundamental for these types of problems.

Question 3. Seg PM and seg PN are congruent chords of a circle with centre C. Show that the ray PC is the bisector of \( \angle \)NPM.
Given: Point C is the centre of the circle.
chord PM \( \simeq \) chord PN
To prove: Ray PC is the bisector of \( \angle \)NPM.
Construction: Draw seg CR \( \perp \) chord PN, P-R-N
seg CQ \( \perp \) chord PM, P-Q-M
Proof:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त का चित्र है जिसका केंद्र C है। इसमें दो सर्वांगसम जीवाएँ PM और PN दर्शाई गई हैं। केंद्र C से जीवा PM पर CQ लंब है और जीवा PN पर CR लंब है।
chord PM \( \simeq \) chord PN [Given]
seg CR \( \perp \) chord PN
seg CQ \( \perp \) chord PM [Construction]
\( \therefore \) segCR = segCQ ....(i) [Congruent chords are equidistant from the centre]
In \( \Delta \)PRC and \( \Delta \)PQC,
\( \angle \)PRC = \( \angle \)PQC [Each is of 90°]
segCR = segCQ [From (i)]
seg PC = seg PC [Common side]
\( \therefore \) \( \Delta \)PRC \( \simeq \) \( \Delta \)PQC [Hypotenuse side test]
\( \therefore \) \( \angle \)RPC \( \simeq \) \( \angle \)QPC [c. a. c. t.]
\( \therefore \) \( \angle \)NPC \( \simeq \) \( \angle \)MPC [N- R-P, M-Q-P]
\( \therefore \) Ray PC is the bisector of \( \angle \)NPM.
Answer: Ray PC is the bisector of \( \angle \)NPM.
In simple words: By drawing perpendiculars from the center to the congruent chords, we establish that the distances from the center to the chords are equal. This allows us to prove the congruence of two right-angled triangles using the Hypotenuse-Side test. Corresponding angles of congruent triangles then show that ray PC bisects the angle formed by the chords.

🎯 Exam Tip: When proving angle bisectors, look for congruent triangles. In circles, using the property that congruent chords are equidistant from the center is a common first step to establish necessary conditions for congruence proofs.

Maharashtra Board Class 9 Maths Chapter 6 Circle Practice Set 6.2 Intext Questions And Activities

Question 1. Prove the following two theorems for two congruent circles. (Textbook pg. no. 81)
(i) Congruent chords in congruent circles are equidistant from their respective centres.
(ii) Chords of congruent circles which are equidistant from their respective centres are congruent.
Write 'Given'. 'To prove' and the proofs of these theorems.
Answer:

(i) Congruent chords in congruent circles are equidistant from their respective centres.
Given: Point P and point Q are the centres of congruent circles.
chord AB \( \simeq \) chord CD
seg PM \( \perp \) chord AB, A-M-B
seg QN \( \perp \) chord CD, C-N-D
To prove: PM = QN
ℹ️ चित्र व्याख्या (Diagram Explanation): ये दो अलग-अलग सर्वांगसम वृत्त हैं। पहले वृत्त का केंद्र P है और इसमें जीवा AB है; PM जीवा AB पर लंब है और उसे M पर समद्विभाजित करता है। PA एक त्रिज्या है। दूसरे वृत्त का केंद्र Q है और इसमें जीवा CD है; QN जीवा CD पर लंब है और उसे N पर समद्विभाजित करता है। QC एक त्रिज्या है।
Construction: Draw seg PA and seg QC.
Proof:
seg PM \( \perp \) chord AB, seg QN \( \perp \) chord CD [Given]
\( \therefore \) AM = \( \frac{1}{2} \)(AB)..........(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
\( \therefore \) CN = \( \frac{1}{2} \)(CD)........(ii) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
But, AB = CD .........(iii) [Given]
\( \therefore \) AM = CN [From (i), (ii) and (iii)]
i.e., segAM \( \simeq \) segCN ....(iv) [Segments of equal lengths]
In \( \Delta \)PMA and \( \Delta \)QNC,
\( \angle \)PMA = \( \angle \)QNC [Each is of 90°]
hypotenuse PA \( \simeq \) hypotenuse QC [Radii of congruent circles]
seg AM = seg CN [From (iv)]
\( \therefore \) \( \Delta \)PMA \( \simeq \) \( \Delta \)QNC [Hypotenuse side test]
\( \therefore \) segPM \( \simeq \) segQN [c. s. c. t.]
\( \therefore \) PM \( \simeq \) QN [Length of congruent segments]

(ii) Chords of congruent circles which are equidistant from their respective centres are congruent.
ℹ️ चित्र व्याख्या (Diagram Explanation): ये दो अलग-अलग सर्वांगसम वृत्त हैं। पहले वृत्त का केंद्र P है और इसमें जीवा AB है; PM जीवा AB पर लंब है और उसे M पर समद्विभाजित करता है। PA एक त्रिज्या है। दूसरे वृत्त का केंद्र Q है और इसमें जीवा CD है; QN जीवा CD पर लंब है और उसे N पर समद्विभाजित करता है। QC एक त्रिज्या है।
Given: Point P and point Q are the centres of congruent circles.
seg PM \( \perp \) chord AB, A-M-B
seg QN \( \perp \) chord CD, C-N-D
PM = QN
To prove: chord AB \( \simeq \) chord CD
Construction: Draw seg PA and seg QC.
Proof:
In \( \Delta \)PMA and \( \Delta \)QNC,
\( \therefore \) \( \angle \)PMA \( \simeq \) \( \angle \)QNC [Each is of 90°]
seg PM = seg QN [Given]
hypotenuse PA \( \simeq \) hypotenuse QC [Radii of the congruent circles]
\( \therefore \) \( \Delta \)PMA \( \simeq \) \( \Delta \)QNC [Hypotenuse side test]
\( \therefore \) seg AM \( \simeq \) seg CN [c. s. c. t.]
\( \therefore \) AM = CN ....(i) [Length of congruent segments]
Now, seg PM \( \perp \) chord AB, and seg QN \( \perp \) chord CD
\( \therefore \) AM = \( \frac{1}{2} \)(AB) ...(ii)
\( \therefore \) CN = \( \frac{1}{2} \)(CD) ..(iii) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
\( \therefore \) AB = CD [From (i), (ii) and (ii)]
\( \therefore \) chord AB \( \simeq \) chord CD [Segments of equal lengths]
Answer:
(i) **Congruent chords in congruent circles are equidistant from their respective centres.**
By drawing perpendiculars from the centers to the chords, these perpendiculars bisect the chords. Since the chords are congruent, their halves are also congruent. Using the Hypotenuse-Side (RHS) test for congruence in the right-angled triangles formed by the radii, half-chords, and perpendicular distances, we prove that the perpendicular distances (distances from the center) are equal.
(ii) **Chords of congruent circles which are equidistant from their respective centres are congruent.**
Given that the chords are equidistant from the centers, we can prove the congruence of the right-angled triangles formed by the radii, half-chords, and perpendicular distances using the Hypotenuse-Side (RHS) test. This establishes that the half-chords are congruent, and consequently, the full chords are also congruent.
In simple words: For two congruent circles, if chords are equal, their distances from the center are equal (Theorem 1). Conversely, if the distances of chords from the center are equal, then the chords themselves are equal (Theorem 2). Both are proven using congruent triangles formed by radii, half-chords, and distances from the center.

🎯 Exam Tip: These two theorems are converses of each other and are fundamental in circle geometry. Mastering their proofs, especially the use of congruent triangles (RHS criterion), is vital for theoretical questions and applying them in problem-solving.