Maharashtra Board Class 10 Maths Chapter 3 Circle Set 3.1 Solutions

Question 1. In the adjoining figure, the radius of a circle with centre C is 6 cm, line AB is a tangent at A. Answer the following questions.
(i) What is the measure of ∠CAB? Why?
(ii) What is the distance of point C from line AB? Why?
(iii) d(A, B) = 6 cm, find d(B, C).
(iv) What is the measure of ∠ABC? Why?

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र C है, और एक रेखा AB वृत्त को बिंदु A पर स्पर्श करती है। बिंदु C वृत्त का केंद्र है और AC वृत्त की त्रिज्या है।
Answer: Solution:
(i) line AB is the tangent to the circle with centre C and radius AC. [Given]
\( \therefore \) ∠CAB = 90° (i) [Tangent theorem]
(ii) seg CA \( \perp \) line AB [From (i)]
radius = I(AC) = 6 cm
\( \therefore \) The distance of point C from line AB is 6 cm.
(iii) In \( \triangle CAB \), ∠CAB = 90° [From (i)]
\( \therefore \) \( BC^2 = AB^2 + AC^2 \). [Pythagoras theorem]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र C है, और एक रेखा AB वृत्त को बिंदु A पर स्पर्श करती है। बिंदु C वृत्त का केंद्र है, AC वृत्त की त्रिज्या है, और एक समकोण त्रिभुज CAB बनता है।
\( = 6^2 + 6^2 \)
\( = 2 \times 6^2 \)
\( \therefore \) \( BC = \sqrt{2 \times 6^2} \) [Taking square root of both sides]
\( = 6\sqrt{2} \) cm
\( \therefore \) d(B, C) = \( 6\sqrt{2} \) cm
(iv) In \( \triangle ABC \),
AC = AB = 6cm
\( \therefore \) ∠ABC = ∠ACB [Isosceles triangle theorem]
Let ∠ABC = ∠ACB =x
In \( \triangle ABC \),
∠CAB + ∠ABC + ∠ACB = 180° [Sum of the measures of angles of a triangle is 180°]
\( \therefore \) 90° + x + x = 180°
\( \therefore \) 90 + 2x = 180°
\( \therefore \) 2x = 180°-90°
\( \therefore \) x = \( \frac{90}{2} \)
\( \therefore \) x = 45°
\( \therefore \) ∠ABC = 45°
In simple words: This problem applies the tangent theorem to find the angle between the radius and tangent, uses the Pythagorean theorem to calculate distances in a right-angled triangle, and applies the isosceles triangle theorem to find angles.

🎯 Exam Tip: Remember to clearly state the theorems used for each step, such as the tangent theorem and Pythagoras theorem, to score full marks.

Question 2. In the adjoining figure, O is the centre of the circle. From point R, seg RM and seg RN are tangent segments touching the circle at M and N. If (OR) = 10 cm and radius of the circle = 5 cm, then
(i) What is the length of each tangent segment?
(ii) What is the measure of ∠MRO?
(iii) What is the measure of ∠MRN?

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र O है। बिंदु R से, दो स्पर्शरेखा खंड RM और RN वृत्त को क्रमशः M और N पर स्पर्श करते हैं। O से M और N तक की रेखाएँ वृत्त की त्रिज्याएँ हैं।
Answer: Solution:
seg RM and seg RN are tangents to the circle with centre O. [Given]
\( \therefore \) ∠OMR = ∠ONR = 90° [Tangent theorem]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र O है। बिंदु R से, दो स्पर्शरेखा खंड RM और RN वृत्त को क्रमशः M और N पर स्पर्श करते हैं। त्रिज्या OM और ON 5 सेमी हैं और OR 10 सेमी है, जिससे समकोण त्रिभुज OMR और ONR बनते हैं।
(i) In \( \triangle OMR \), ∠OMR = 90°
\( \therefore \) \( OR^2 = OM^2 + RM^2 \) [Pythagoras theorem]
\( \therefore \) \( 10^2 = 5^2 + RM^2 \)
\( \therefore \) 100 = 25 + \( RM^2 \)
\( \therefore \) \( RM^2 \) = 75
\( \therefore \) \( RM = \sqrt{75} \) [Taking square root of both sides]
\( \therefore \) \( RM = 5\sqrt{3} \) cm
\( \therefore \) RM = RN [Tangent segment theorem]
Length of each tangent segment is \( 5\sqrt{3} \) cm.
(ii) In \( \triangle RMO \),
∠OMR = 90° [Tangent theorem]
OM = 5 cm and OR = 10 cm
\( \therefore \) OM = \( \frac{1}{2} \) OR
\( \therefore \) ∠MRO = 30° (i) [Converse of 30-60-90 theorem]
Similarly, ∠NRO = 30°
(iii) But, ∠MRN = ∠MRO + ∠NRO [Angle addition property]
= 30° + 30° [From (i)]
\( \therefore \) ∠MRN = 60°
In simple words: This problem uses the tangent theorem to establish right angles, the Pythagorean theorem to find the length of tangent segments, and the properties of 30-60-90 triangles and angle addition to find the angles.

🎯 Exam Tip: When using the converse of the 30-60-90 theorem, ensure the side ratios (e.g., opposite 30° is half the hypotenuse) are clearly demonstrated. Also, remember tangent segments from an external point are equal in length.

Question 3. Seg RM and seg RN are tangent segments of a circle with centre O. Prove that seg OR bisects ∠MRN as well as ∠MON.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र O है। बिंदु R से, दो स्पर्शरेखा खंड RM और RN वृत्त को क्रमशः M और N पर स्पर्श करते हैं। रेखा OR वृत्त के केंद्र को बाहरी बिंदु R से जोड़ती है।
Answer: Solution:
Proof:
In \( \triangle OMR \) and \( \triangle ONR \),
seg RM \( \cong \) seg RN [Tangent segment theorem]
seg OM \( \cong \) seg ON [Radii of the same circle]
seg OR \( \cong \) seg OR [Common side]
\( \therefore \triangle OMR \cong \triangle ONR \) [SSS test of congruency]
\( \therefore \) { ∠MRO \( \cong \) ∠NRO
∠MOR \( \cong \) ∠NOR } [c.a.c.t.]
\( \therefore \) seg OR bisects ∠MRN and ∠MON.
In simple words: By proving the two triangles formed by the tangents, radii, and the segment from the center to the external point are congruent using SSS test, we can show that the segment from the center bisects both the angle between the tangents and the angle at the center.

🎯 Exam Tip: For proving congruence, clearly identify the corresponding sides and angles and state the congruence test (e.g., SSS, SAS, ASA, AAS, Hypotenuse-Side) correctly. Remember c.a.c.t. means corresponding angles of congruent triangles.

Question 4. What is the distance between two parallel tangents of a circle having radius 4.5 cm? Justify your answer.

Answer: Solution:
Let the lines PQ and RS be the two parallel tangents to circle at M and N respectively.
Through centre O, draw line AB || line RS.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र O है। दो समानांतर स्पर्शरेखाएँ PQ और RS वृत्त को क्रमशः M और N पर स्पर्श करती हैं। एक रेखा AB केंद्र O से होकर खींची गई है जो स्पर्शरेखाओं के समानांतर है और M और N से होकर गुजरती है।
OM = ON = 4.5 [Given]
line AB || line RS [Construction]
line PQ || line RS [Given]
\( \therefore \) line AB || line PQ || line RS
Now, ∠OMP = ∠ONR = 90° (i) [Tangent theorem]
For line PQ || line AB,
∠OMP = ∠AON = 90° (ii) [Corresponding angles and from (i)]
For line RS || line AB,
∠ONR = ∠AOM = 90° (iii) [Corresponding angles and from (i)]
\( \therefore \) ∠AON + ∠AOM = 90° + 90° [From (ii) and (iii)]
\( \therefore \) ∠AON + ∠AOM = 180°
\( \therefore \) ∠AON and ∠AOM form a linear pair.
\( \therefore \) ray OM and ray ON are opposite rays.
\( \therefore \) Points M, O, N are collinear. (iv)
\( \therefore \) MN = OM + ON [M - O - N, From (iv)]
\( \therefore \) MN = 4.5 + 4.5
\( \therefore \) MN = 9 cm
\( \therefore \) Distance between two parallel tangents PQ and RS is 9 cm.
In simple words: The distance between two parallel tangents of a circle is equal to the diameter of the circle. Since the radius is 4.5 cm, the diameter is 9 cm, which is the required distance.

🎯 Exam Tip: Remember that the line joining the points of contact of two parallel tangents always passes through the center of the circle and is perpendicular to both tangents, thus forming the diameter.

Question 1. In the adjoining figure, seg QR is a chord of the circle with centre O. P is the midpoint of the chord QR. If QR = 24, OP = 10, find radius of the circle. To find solution of the problem, write the theorems that are useful. Using them, solve the problem. (Textbook pg. no. 48)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र O है। QR वृत्त की एक जीवा है और P जीवा QR का मध्यबिंदु है। OP जीवा QR पर लंब है, और OQ वृत्त की त्रिज्या है।
Answer: Solution:
Theorems which are useful to find solution:
(i) The segment joining the centre of a circle and the midpoint of a chord is perpendicular to the chord.
(ii) In a right angled triangle, sum of the squares of the perpendicular sides is equal to square of its hypotenuse.
QP = \( \frac{1}{2} \) (QR) [P is the midpoint of chord QR]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र O है। QR जीवा है, P इसका मध्यबिंदु है। OP 10 इकाई है और QR 24 इकाई है। त्रिभुज OPQ एक समकोण त्रिभुज है जहाँ OQ त्रिज्या है।
\( \frac{1}{2} \times 24 \) = 12 units
Also, seg OP \( \perp \) chord QR [The segment joining centre of a circle and midpoint of a chord is perpendicular to the chord]
In \( \triangle OPQ \), ∠OPQ = 90°
\( \therefore \) \( OQ^2 = OP^2 + QP^2 \) [Pythagoras theorem]
\( = 10^2 + 12^2 \)
\( = 100 + 144 \)
\( = 244 \)
\( \therefore \) \( OQ = \sqrt{244} = 2\sqrt{61} \) units.
\( \therefore \) The radius of the circle is \( 2\sqrt{61} \) units.
In simple words: To find the radius, we first use the property that a line from the center to the midpoint of a chord is perpendicular to the chord. Then, we apply the Pythagorean theorem to the right-angled triangle formed to find the hypotenuse, which is the radius.

🎯 Exam Tip: Clearly state the two main theorems used: the perpendicularity of the line from the center to a chord's midpoint, and the Pythagorean theorem. Show all calculations for the segments and the final radius.

Question 2. In the adjoining figure, M is the centre of the circle and seg AB is a diameter, seg MS \( \perp \) chord AD, seg MT \( \perp \) chord AC, ∠DAB = ∠CAB.
(i) Prove that: chord AD \( \cong \) chord AC.
(ii) To solve this problem which theorems will you use?
a. The chords which are equidistant from the centre are equal in length.
b. Congruent chords of a circle are equidistant from the centre.
(iii) Which of the following tests of congruence of triangles will be useful?
a. SAS b. ASA c. SSS d. AAS e. Hypotenuse-side test.
Using appropriate test and theorem write the proof of the above example. (Textbook pg. no, 48)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र M है और AB इसका व्यास है। MS जीवा AD पर लंब है और MT जीवा AC पर लंब है। ∠DAB और ∠CAB समान कोण हैं, जो वृत्त के अंदर दो त्रिभुज MAD और MAC बनाते हैं।
Answer: Solution:
Proof:
(i) ∠DAB = ∠CAB [Given]
\( \therefore \) ∠SAM = ∠TAM (i) [A - S - D, A - M - B, A - T - C]
In \( \triangle SAM \) and \( \triangle TAM \),
∠SAM = ∠TAM [From (i)]
∠ASM = ∠ATM [Each angle is of measure 90°]
seg AM = seg AM [Common side]
\( \therefore \triangle SAM \cong \triangle TAM \) [AAS [SAA] test of congruency]
\( \therefore \) side MS \( \cong \) side MT [c.s.c.t]
But, seg MS \( \perp \) chord AD [Given]
seg MT \( \perp \) chord AC
\( \therefore \) chord AD \( \cong \) chord AC [Chords of a circle equidistant from the centre are congruent]
(ii) Theorem used for solving the problem:
The chords which are equidistant from the centre are equal in length.
(iii) Test of congruency useful in solving the above problem is AAS ISAAI test of congruency.
In simple words: By proving two triangles (SAM and TAM) congruent using the AAS test, we show that the perpendicular distances from the center to the chords (MS and MT) are equal. This implies that the chords AD and AC themselves are equal in length, based on the theorem that chords equidistant from the center are congruent.

🎯 Exam Tip: Clearly state the congruence criterion (AAS in this case) and the theorem (chords equidistant from the center are equal) used in your proof. Ensure all given information is correctly applied.

Question 3.
(i) Draw segment AB. Draw perpendicular bisector I of the segment AB. Take point P on the line I as centre, PA as radius and draw a circle. Observe that the circle passes through point B also. Find the reason.
(ii) Taking any other point Q on the line I, if a circle is drawn with centre Q and radius QA, will it pass through B? Think.
(iii) How many such circles can be drawn, passing through A and B? Where will their centres lie? (Textbook pg. no. 49)

Answer: Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक रेखाखंड AB और उसकी लंब समद्विभाजक रेखा 'l' को दर्शाता है। P और Q रेखा 'l' पर दो बिंदु हैं। P को केंद्र मानकर PA त्रिज्या का एक वृत्त खींचा गया है, और Q को केंद्र मानकर QA त्रिज्या का एक और वृत्त खींचा गया है। दोनों वृत्त B से होकर गुजरते हैं।
(i) Draw the circle with centre P and radius PA.
line I is the perpendicular bisector of seg AB.
Every point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.
\( \therefore \) PA = PB ... [Perpendicular bisector theorem]
\( \therefore \) PA = PB = radius
\( \therefore \) The circle with centre P and radius PA passes through point B.
(ii) The circle with any other point Q and radius QA is drawn.
QA = QB = radius ... [Perpendicular bisector theorem]
\( \therefore \) The circle with centre Q and radius QA passes through point B.
(iii) We can draw infinite number of circles passing through A and B.
All their centres will lie on the perpendicular bisector of AB (i.e., line I)
In simple words: Any point on the perpendicular bisector of a segment is equidistant from the segment's endpoints. Thus, if a circle's center lies on this bisector and its radius is the distance to one endpoint, it will also pass through the other endpoint. An infinite number of such circles can be drawn, with their centers always on the perpendicular bisector.

🎯 Exam Tip: The key concept here is the perpendicular bisector theorem. Emphasize that all points equidistant from two given points lie on the perpendicular bisector of the segment connecting them. This is crucial for understanding circle centers.

Question 4.
(i) Take any three non-collinear points. What should be done to draw a circle passing through all these points? Draw a circle through these points.
(ii) Is it possible to draw one more circle passing through these three points? Think of it. (Textbook pg. no. 49)

Answer: Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह तीन असंरेख बिंदुओं A, B, C को दर्शाता है। रेखाखंड AB और BC के लंब समद्विभाजक (क्रमशः रेखा 'l' और 'm') बिंदु P पर प्रतिच्छेद करते हैं। P को केंद्र मानकर, PA त्रिज्या का एक वृत्त खींचा गया है, जो A, B और C तीनों बिंदुओं से होकर गुजरता है।
(i) Let points A, B, C be any three non collinear points.
Draw the perpendicular bisector of seg AB (line I).
\( \therefore \) Points A and B are equidistant from any point of line I ....(i) [Perpendicular bisector theorem]
Draw the perpendicular bisector of seg BC (line m) to intersect line I at point P.
\( \therefore \) Points B and C are equidistant from any point of line m ....(ii) [Perpendicular bisector theorem]
\( \therefore \) PA = PB ...[From (i)]
PB = PC ... [From (ii)]
\( \therefore \) PA = PB = PC = radius
\( \therefore \) With PA as radius the required circle is drawn through points A, B, C.
(ii) It is not possible to draw more than one circle passing through these three points.
In simple words: To draw a circle through three non-collinear points, find the intersection of the perpendicular bisectors of any two segments formed by these points. This intersection point is the unique center of the circle, and the distance to any of the three points is its radius. Only one such unique circle can be drawn.

🎯 Exam Tip: For problems involving three non-collinear points, remember that their circumcenter (the center of the circumscribed circle) is the intersection of the perpendicular bisectors of the sides of the triangle formed by these points. This center is unique.

Question 5. Take 3 collinear points D, E, F. Try to draw a circle passing through these points. If you are not able to draw a circle, think of the reason. (Textbook pg. no. 49)

Answer: Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह तीन संरेख बिंदुओं D, E, F को दर्शाता है। रेखाखंड DE और EF के लंब समद्विभाजक (क्रमशः रेखा 'l' और 'm') समानांतर हैं और कभी प्रतिच्छेद नहीं करते हैं।
Let D, E, F be the collinear points.
The perpendicular bisector of DE and EF drawn (i.e., line I and line m) do not intersect at a common point.
\( \therefore \) There is no single common point (centre of circle) from which a circle can be drawn passing through points D, E and F.
Hence, we cannot draw a circle passing through points D, E and F.
In simple words: It's impossible to draw a circle through three collinear points because the perpendicular bisectors of the segments formed by these points would be parallel, meaning they would never intersect to form a common center for the circle. A circle requires its center to be equidistant from all points on its circumference, which cannot be achieved with collinear points.

🎯 Exam Tip: Understand that for a circle to pass through multiple points, its center must be equidistant from all those points. For collinear points, this condition cannot be met as their perpendicular bisectors are parallel and never intersect, thus no unique center exists.

Question 6. Which theorem do we use in proving that hypotenuse is the longest side of a right angled triangle? (Textbook pg. no. 52)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज ABC को दर्शाता है, जिसमें कोण B 90° का है। AC इस त्रिभुज का कर्ण है, और AB तथा BC भुजाएँ हैं।
Answer: Solution:
In \( \triangle ABC \),
∠ABC = 90°
\( \therefore \) ∠BAC < 90° and ∠ACB < 90° [Given]
\( \therefore \) ∠ABC > ∠BAC and ∠ABC > ∠ACB
\( \therefore \) AC > BC and AC > AB [Side opposite to greater angle is greater]
\( \therefore \) Hypotenuse is the longest side in right angled triangle.
We use theorem, If two angles of a triangle are not equal, then the side opposite to the greater angle is greater than the side opposite to the smaller angle.
In simple words: The theorem used states that in any triangle, the side opposite the larger angle is longer than the side opposite the smaller angle. In a right-angled triangle, the right angle (90°) is always the largest angle, so the side opposite it, the hypotenuse, must be the longest side.

🎯 Exam Tip: To prove the hypotenuse is the longest side, first establish that the right angle is the largest angle in the triangle (since the sum of angles is 180° and other angles are acute). Then, directly apply the theorem about the relationship between angles and opposite sides.

Question 7. Theorem: Tangent segments drawn from an external point to a circle are congruent Draw radius AP and radius AQ and complete the following proof of the theorem. Given: A is the centre of the circle. Tangents through external point D touch the circle at the points P and Q. To prove: seg DP \( \cong \) seg DQ Construction: Draw seg AP and seg AQ.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र A है। बिंदु D से, दो स्पर्शरेखाएँ वृत्त को क्रमशः P और Q पर स्पर्श करती हैं। AP और AQ वृत्त की त्रिज्याएँ हैं, जो स्पर्शरेखाओं DP और DQ से मिलती हैं।
Answer: Solution:
Proof:
In \( \triangle PAD \) and \( \triangle QAD \),
seg PA \( \cong \) [segQA] [Radii of the same circle]
seg AD \( \cong \) seg AD [Common side]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र A है। बिंदु D से, दो स्पर्शरेखाएँ वृत्त को क्रमशः P और Q पर स्पर्श करती हैं। AP और AQ वृत्त की त्रिज्याएँ हैं। त्रिभुज APD और AQD समकोण त्रिभुज हैं, जहाँ AP और AQ त्रिज्याएँ हैं और AD उभयनिष्ठ भुजा है।
∠APD = ∠AQD = 90° [Tangent theorem]
\( \therefore \triangle PAD \cong \triangle QAD \) [By Hypotenuse side test]
\( \therefore \) seg DP \( \cong \) seg DQ [c.s.c.t]
In simple words: This proof establishes that two triangles formed by the center, the external point, and the tangent points are congruent using the Hypotenuse-Side (RHS) test. Since the triangles are congruent, their corresponding sides, which are the tangent segments, must also be congruent.

🎯 Exam Tip: When proving tangent segment congruence, the Hypotenuse-Side (RHS) congruence test is often used because the radii are perpendicular to the tangents at the point of contact, creating right-angled triangles with a common hypotenuse.