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

Question 1. Choose correct alternative answer and fill in the blanks.

(i) Radius of a circle is 10 cm and distance of a chord from the centre is 6 cm. Hence, the length of the chord is _______.
(A) 16 cm
(B) 8 cm
(C) 12 cm
(D) 32 cm
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र O है। एक जीवा AB वृत्त के अंदर है, और केंद्र O से जीवा AB पर एक लम्ब खींचा गया है, जो AB को C पर काटता है। OA वृत्त की त्रिज्या है, और OC केंद्र से जीवा की दूरी है। यह एक समकोण त्रिभुज OAC बनाता है।
\( \therefore OA^2 = AC^2 + OC^2 \)
\( \therefore 10^2 = AC^2 + 6^2 \)
\( \therefore AC^2 = 64 \)
\( \therefore AC = 8 \text{ cm} \)
\( \therefore AB = 2(AC) = 16 \text{ cm} \)
(A) 16 cm
In simple words: Using the Pythagorean theorem, with the radius as the hypotenuse and the distance from the center to the chord as one leg, we find half the chord length. Doubling this gives the full length of the chord.

🎯 Exam Tip: Remember that a perpendicular from the center to a chord bisects the chord. This is a fundamental property used in many circle geometry problems.

Question 1.
(ii) The point of concurrence of all angle bisectors of a triangle is called the _______.
(A) centroid
(B) circumcentre
(C) incentre
(D) orthocentre
Answer:
(C) incentre
In simple words: The incenter is the point where the angle bisectors of a triangle meet, and it's the center of the triangle's incircle.

🎯 Exam Tip: Understanding the definitions of centroid, circumcenter, incenter, and orthocenter is crucial. Each represents the intersection of different types of lines within a triangle.

Question 1.
(iii) The circle which passes through all the vertices of a triangle is called _______.
(A) circumcircle
(B) incircle
(C) congruent circle
(D) concentric circle
Answer:
(A) circumcircle
In simple words: A circumcircle is a circle that passes through all three vertices of a triangle. Its center is called the circumcenter.

🎯 Exam Tip: Distinguish between incircle (touches sides internally) and circumcircle (passes through vertices). Their properties and centers are distinct.

Question 1.
(iv) Length of a chord of a circle is 24 cm. If distance of the chord from the centre is 5 cm, then the radius of that circle is _______.
(A) 12 cm
(B) 13 cm
(C) 14 cm
(D) 15 cm
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र O है। एक जीवा AB वृत्त के अंदर है। केंद्र O से जीवा AB पर एक लम्ब खींचा गया है, जो AB को C पर काटता है। OA वृत्त की त्रिज्या है, और OC केंद्र से जीवा की दूरी है। जीवा की आधी लंबाई AC है।
\( OA^2 = AC^2 + OC^2 \)
\( \therefore OA^2 = 12^2 + 5^2 \)
\( \therefore OA^2 = 169 \)
\( \therefore OA = 13 \text{ cm} \)
(B) 13 cm
In simple words: Given the chord length, we find half its length. Then, using the distance from the center and half the chord length with the Pythagorean theorem, we can calculate the radius.

🎯 Exam Tip: Always draw a clear diagram for such problems. Visualizing the right-angled triangle formed by the radius, half-chord, and distance from the center is key to applying Pythagoras' theorem correctly.

Question 1.
(v) The length of the longest chord of the circle with radius 2.9 cm is _______.
(A) 3.5 cm
(B) 7 cm
(C) 10 cm
(D) 5.8 cm
Answer:
Longest chord of the circle = diameter = 2 x radius = 2 x 2.9 = 5.8 cm
(D) 5.8 cm
In simple words: The longest chord in any circle is its diameter, which is always twice the radius.

🎯 Exam Tip: This is a fundamental definition in circles. Clearly stating that the longest chord is the diameter will earn full marks.

Question 1.
(vi) Radius of a circle with centre O is 4 cm. If l(OP) = 4.2 cm, say where point P will lie _______.
(A) on the centre
(B) inside the circle
(C) outside the circle
(D) on the circle
Answer:
l(OP) > radius
\( \therefore \) Point P lies in the exterior of the circle.
(C) outside the circle
In simple words: If the distance of a point from the center is greater than the radius, the point lies outside the circle.

🎯 Exam Tip: Remember the relationship between the distance from the center to a point (d) and the radius (r): if d < r, point is inside; if d = r, point is on the circle; if d > r, point is outside.

Question 1.
(vii) The lengths of parallel chords which are on opposite sides of the centre of a circle are 6 cm and 8 cm. If radius of the circle is 5 cm, then the distance between these chords is _______.
(A) 2 cm
(B) 1 cm
(C) 8 cm
(D) 7 cm
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक वृत्त है जिसका केंद्र O है। दो समांतर जीवाएं PQ और MN केंद्र के विपरीत दिशाओं में हैं। केंद्र O से जीवा PQ पर लंब AQ है, और केंद्र O से जीवा MN पर लंब BN है। OA और OB वृत्त की त्रिज्याएं हैं। AB इन दो जीवाओं के बीच की दूरी है।
\( PQ = 8 \text{ cm, } MN = 6 \text{ cm} \)
\( \therefore AQ = 4 \text{ cm, } BN = 3 \text{ cm} \)
\( \therefore OQ^2 = OA^2 + AQ^2 \)
\( \therefore 5^2 = OA^2 + 4^2 \)
\( \therefore OA^2 = 25 - 16 = 9 \)
\( \therefore OA = 3 \text{ cm} \)
Also, \( ON^2 = OB^2 + BN^2 \)
\( \therefore 5^2 = OB^2 + 3^2 \)
\( \therefore OB^2 = 25 - 9 = 16 \)
\( \therefore OB = 4 \text{ cm} \)
Now, \( AB = OA + OB = 3 + 4 = 7 \text{ cm} \)
In simple words: For each chord, half its length and the radius form a right-angled triangle with the distance from the center. Calculate each distance from the center using the Pythagorean theorem, then add them since the chords are on opposite sides.

🎯 Exam Tip: When chords are on opposite sides of the center, their distances from the center are added to find the total distance between them. If they were on the same side, the distances would be subtracted.

Question 2. Construct incircle and circumcircle of an equilateral ADSP with side 7.5 cm. Measure the radii of both the circles and find the ratio of radius of circumcircle to the radius of incircle.
Answer: 
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समबाहु त्रिभुज DPS को दर्शाता है जिसकी प्रत्येक भुजा 7.5 cm है। त्रिभुज के अंदर एक छोटा वृत्त (अंतर्वृत्त) है जो उसकी सभी भुजाओं को स्पर्श करता है। त्रिभुज के बाहर एक बड़ा वृत्त (परिवृत्त) है जो उसके तीनों शीर्षों से होकर गुजरता है। बिंदु C त्रिभुज का केंद्र है, जो अंतर्वृत्त और परिवृत्त दोनों का केंद्र भी है।
Steps of construction:
(i) Construct \( \triangle \)DPS of the given measurement.
(ii) Draw the perpendicular bisectors of side DP and side PS of the triangle.
(iii) Name the point of intersection of the perpendicular bisectors as point C.
(iv) With C as centre and CM as radius, draw a circle which touches all the three sides of the triangle.
(v) With C as centre and CP as radius, draw a circle which passes through the three vertices of the triangle.
Radius of incircle = 2.2 cm and Radius of circumcircle = 4.4 cm
\( \frac{\text{Radius of circumcircle}}{\text{Radius of incircle}} = \frac{4.4}{2.2} \)
\( = \frac{44}{22} \)
\( = \frac{2}{1} \)
\( = 2:1 \)
In simple words: First, construct the equilateral triangle. Then, draw perpendicular bisectors to find the center (which is common for both incircle and circumcircle in an equilateral triangle). Use the distance from the center to a side for the incircle radius and to a vertex for the circumcircle radius, then calculate their ratio.

🎯 Exam Tip: For an equilateral triangle, the incenter, circumcenter, centroid, and orthocenter all coincide. The ratio of circumradius to inradius in an equilateral triangle is always 2:1.

Question 3. Construct \( \triangle \)NTS where NT = 5.7 cm. TS = 7.5 cm and \( \angle \)NTS = 110° and draw incircle and circumcircle of it.
Answer: 
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज NTS को दर्शाता है जिसमें भुजा NT 5.7 cm, भुजा TS 7.5 cm और कोण NTS 110° है। त्रिभुज के अंदर एक अंतर्वृत्त (incircle) है जो त्रिभुज की सभी भुजाओं को स्पर्श करता है, और त्रिभुज के बाहर एक परिवृत्त (circumcircle) है जो त्रिभुज के तीनों शीर्षों से होकर गुजरता है। अंतर्वृत्त का केंद्र I है और परिवृत्त का केंद्र C है।
Steps of construction:
For incircle:
(i) Construct \( \triangle \)NTS of the given measurement.
(ii) Draw the bisectors of \( \angle \)T and \( \angle \)S. Let these bisectors intersect at point I.
(iii) Draw a perpendicular IM on side TS. Point M is the foot of the perpendicular.
(iv) With I as centre and IM as radius, draw a circle which touches all the three sides of the triangle.
For circumcircle:
(i) Draw the perpendicular bisectors of side NT and side TS of the triangle.
(ii) Name the point of intersection of the perpendicular bisectors as point C.
(iii) Join seg CN
(iv) With C as centre and CN as radius, draw a circle which passes through the three vertices of the triangle.
In simple words: To draw the incircle, construct the triangle, find the incenter by bisecting two angles, drop a perpendicular to a side for the radius, then draw the circle. For the circumcircle, construct the triangle, find the circumcenter by drawing perpendicular bisectors of two sides, use the distance to a vertex for the radius, then draw the circle.

🎯 Exam Tip: For incircles, bisect angles to find the incenter; for circumcircles, bisect sides to find the circumcenter. Practice constructing perpendicular bisectors and angle bisectors accurately for precise results.

Question 4. In the adjoining figure, C is the centre of the circle, seg QT is a diameter, CT = 13, CP = 5. Find the length of chord RS.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक वृत्त है जिसका केंद्र C है। QT वृत्त का व्यास है। RS वृत्त की एक जीवा है, और CP केंद्र C से जीवा RS पर एक लंब है, जो P पर जीवा को काटता है। CT वृत्त की त्रिज्या है।
Given: In a circle with centre C, QT is a diameter, CT = 13 units, CP = 5 units
To find: Length of chord RS
Construction: Join points R and C.
Answer: 

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र C है। RS वृत्त की एक जीवा है। CT वृत्त की त्रिज्या है। CP जीवा RS पर लंब है, P पर काटता है। CR को जोड़ा गया है, जो वृत्त की त्रिज्या है। त्रिभुज CPR एक समकोण त्रिभुज है।
(i) \( CR = CT = 13 \) units .....(i) [Radii of the same circle]
In \( \triangle \)CPR, \( \angle CPR = 90^\circ \)
\( \therefore CR^2 = CP^2 + RP^2 \) [Pythagoras theorem]
\( \therefore 13^2 = 5^2 + RP^2 \) [From (i)]
\( \therefore 169 = 25 + RP^2 \) [From (i)]
\( \therefore RP^2 = 169 - 25 = 144 \)
\( \therefore RP = \sqrt{144} \) [Taking square root on both sides]
\( \therefore RP = 12 \text{ cm} \) ....(ii)
(ii) Now, seg CP \( \perp \) chord RS [Given]
\( \therefore RP = \frac{1}{2} RS \) [Perpendicular drawn from the centre of the circle to the chord
bisects the chord.]
\( \therefore 12 = \frac{1}{2} RS \) [From (ii)]
\( \therefore RS = 2 \times 12 = 24 \)
\( \therefore \) The length of chord RS is 24 units.
In simple words: First, recognize that CR is a radius, so CR = CT. Then, in the right-angled triangle CPR, use Pythagoras' theorem to find RP (half of the chord RS). Double RP to find the full length of chord RS.

🎯 Exam Tip: Always remember that the perpendicular from the center of a circle to a chord bisects the chord. This property, combined with Pythagoras' theorem, is essential for solving such problems.

Question 5. In the adjoining figure, P is the centre of the circle. Chord AB and chord CD intersect on the diameter at the point E. If \( \angle AEP \cong \angle DEP \), then prove that AB = CD.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र P है। वृत्त के अंदर दो जीवाएं AB और CD हैं जो व्यास पर बिंदु E पर प्रतिच्छेद करती हैं। कोण AEP और DEP बराबर दिखाए गए हैं, जो दर्शाते हैं कि PE कोण AED का समद्विभाजक है।
Given: P is the centre of the circle.
Chord AB and chord CD intersect on the diameter at the point E. \( \angle AEP \cong \angle DEP \)
To prove: AB = CD
Construction: Draw seg PM \( \perp \) chord AB, A-M-B
seg PN \( \perp \) chord CD, C-N-D
Proof:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में वृत्त का केंद्र P है। जीवाएं AB और CD बिंदु E पर प्रतिच्छेद करती हैं। PM जीवा AB पर लंब है और PN जीवा CD पर लंब है। PE कोण AED का समद्विभाजक है।
\( \angle AEP \cong \angle DEP \) [Given]
\( \therefore \) Seg ES is the bisector of \( \angle \)AED.
Point P is on the bisector of \( \angle \)AED.
\( \therefore PM = PN \) [Every point on the bisector of an angle is equidistant from the sides of the angle.]
\( \therefore \) chord AB = chord CD [Chords which are equidistant from the centre are congruent.]
\( \therefore AB = CD \) [Length of congruent segments]
In simple words: If the diameter bisects the angle formed by two intersecting chords, it means the center is equidistant from those chords. Chords that are equidistant from the center of a circle are equal in length.

🎯 Exam Tip: The key theorem here is that chords equidistant from the center are congruent. If you can prove that PM = PN (distances from the center to the chords), then AB = CD directly follows.

Question 6. In the adjoining figure, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that \( \triangle \)ABC is an isosceles triangle.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र O है। CD वृत्त का व्यास है। AB वृत्त की एक जीवा है। व्यास CD जीवा AB पर बिंदु E पर लंब है। त्रिभुज ABC को बनाया गया है।
Given: O is the centre of the circle.
diameter CD \( \perp \) chord AB, A-E-B
To prove: \( \triangle \)ABC is an isosceles triangle.
Proof:
diameter CD \( \perp \) chord AB [Given]
\( \therefore \) seg OE \( \perp \) chord AB [C-O-E, O-E-D]
\( \therefore \) seg AE \( \cong \) seg BE ......(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
In \( \triangle \)CEA and \( \triangle \)CEB,
\( \angle CEA = \angle CEB \) [Each is of \( 90^\circ \)]
seg AE = seg BE [From (i)]
seg CE = seg CE [Common side]
\( \therefore \triangle CEA \cong \triangle CEB \) [SAS test]
\( \therefore \) seg AC = seg BC [c. s. c. t.]
\( \therefore \triangle ABC \) is an isosceles triangle.
In simple words: Since the diameter is perpendicular to the chord AB, it bisects AB, making AE = BE. Then, by comparing triangles CEA and CEB using SAS congruence, we prove that AC = BC, which means triangle ABC is isosceles.

🎯 Exam Tip: The property that a perpendicular from the center bisects a chord is crucial. Use this to establish congruent segments and then apply triangle congruence criteria (like SAS) to prove the isosceles nature.

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

Question 1. Every student in the group should do this activity. Draw a circle in your notebook. Draw any chord of that circle. Draw perpendicular to the chord through the centre of the circle. Measure the lengths of the two parts of the chord. Group leader should prepare a table as shown below and ask other students to write their observations in it. Write the property which you have observed. (Textbook pg. no. 77)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र O है। वृत्त के अंदर एक जीवा AB है। केंद्र O से जीवा AB पर एक लम्ब OP खींचा गया है, जो जीवा को P पर काटता है। यह दर्शाता है कि जीवा के दो भाग AP और PB हैं।

Answer:
On completing the above table, you will observe that the perpendicular drawn from the centre of a circle on its chord bisects the chord.
In simple words: When you draw a perpendicular from the center of a circle to any chord, it divides the chord into two equal parts.

🎯 Exam Tip: This is a fundamental property of circles. Make sure to accurately draw perpendiculars and measure to confirm this observation.

Question 2. Every student from the group should do this activity. Draw a circle in your notebook. Draw a chord of the circle. Join the midpoint of the chord and centre of the circle. Measure the angles made by the segment with the chord. Discuss about the measures of the angles with your friends. Which property do the observations suggest ? (Textbook pg. no. 77)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त को दर्शाता है जिसका केंद्र है। वृत्त के अंदर एक जीवा है। जीवा के मध्यबिंदु को वृत्त के केंद्र से एक रेखाखंड द्वारा जोड़ा गया है।
Answer:
The measure of the angles made by the drawn segment with the chord is \( 90^\circ \).
Thus, we can conclude that, the segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord.
In simple words: If you connect the center of a circle to the midpoint of a chord, the connecting line segment will always be perpendicular to that chord.

🎯 Exam Tip: This is the converse of the previous property. Both are important and often used together in problems. Precise drawing and measurement will reinforce this concept.

Question 3. Draw circles of convenient radii. Draw two equal chords in each circle. Draw perpendicular to each chord from the centre. Measure the distance of each chord from the centre. What do you observe? (Textbook pg. no. 79)
Answer: 
Congruent chords of a circle are equidistant from the centre.
In simple words: In any circle, if two chords have the same length, then their perpendicular distances from the center of the circle will also be equal.

🎯 Exam Tip: This property is crucial for proving the equality of chords or distances. Accuracy in drawing equal chords and measuring their distances from the center is vital for this activity.

Question 4. Measure the lengths of the perpendiculars on chords in the following figures.
ℹ️ चित्र व्याख्या (Diagram Explanation):
Figure (i): एक वृत्त है जिसका केंद्र O है। जीवा LM है और केंद्र O से जीवा LM पर लंब OL खींचा गया है।
Figure (ii): एक वृत्त है जिसका केंद्र P है। जीवा NT है और केंद्र P से जीवा NT पर लंब PN खींचा गया है।
Figure (iii): एक वृत्त है जिसका केंद्र B है। जीवा MA है और केंद्र B से जीवा MA पर लंब MB खींचा गया है।
Did you find OL = OM in fig (i), PN = PT in fig (ii) and MA = MB in fig (iii)?
Write the property which you have noticed from this activity. (Textbook pg. no. 80)
Answer:
In each figure, the chords are equidistant from the centre. Also, we can see that the measures of the chords in each circle are equal.
Thus, we can conclude that chords of a circle equidistant from the centre of a circle are congruent.
In simple words: This activity reinforces that chords that are the same distance from the center of a circle must have the same length.

🎯 Exam Tip: This is the converse of the property from Question 3. Both "chords equidistant from the center are congruent" and "congruent chords are equidistant from the center" are important theorems.

Question 5. Draw different triangles of different measures and draw in circles and circumcircles of them. Complete the table of observations and discuss. (Textbook pg. no. 85)
Answer: 

In simple words: The incenter is always inside any triangle. The circumcenter's position varies: inside for acute, on the hypotenuse for right-angled, and outside for obtuse triangles.

🎯 Exam Tip: Memorize the location of the incenter and circumcenter for different types of triangles. This knowledge is crucial for understanding geometric properties and for constructions. The incenter is always inside.