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Circles

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Summary

Chapter Summary: Circles

Key Points

  • A circle is defined as a collection of all points in a plane at a constant distance (radius) from a fixed point (centre).
  • Types of Lines with Respect to a Circle:
    • Non-Intersecting Line: No common points with the circle.
    • Secant: Two common points with the circle.
    • Tangent: One common point with the circle.

Important Properties

  • The tangent to a circle is perpendicular to the radius at the point of contact.
  • The lengths of the two tangents drawn from an external point to a circle are equal.

Summary of Activities

  1. Tangent Existence:
    • No tangent can be drawn from a point inside the circle.
    • One tangent can be drawn from a point on the circle.
    • Two tangents can be drawn from a point outside the circle.

Theorems

  • Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Exercises

  • Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
  • Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
  • Find the radius of a circle given the length of a tangent and the distance from the centre.
  • Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Learning Objectives

  • Understand the concept of tangents to a circle.
  • Identify the properties of tangents, including their relationship with the radius.
  • Prove that the lengths of tangents from an external point to a circle are equal.
  • Demonstrate the number of tangents that can be drawn from various positions relative to a circle.
  • Apply theorems related to tangents and circles in problem-solving.

Detailed Notes

Chapter 10: Circles

10.1 Introduction

  • A circle is a collection of all points in a plane at a constant distance (radius) from a fixed point (centre).
  • Key terms related to circles: chord, segment, sector, arc.

10.2 Tangent to a Circle

  • A tangent intersects the circle at only one point.
  • Activities to Understand Tangents:
    • Activity 1: Rotate a straight wire around a point on a circular wire to observe tangent behavior.
    • Case 1: No tangent exists for points inside the circle.
    • Case 2: One tangent exists for points on the circle.
    • Case 3: Two tangents exist for points outside the circle.

10.3 Number of Tangents from a Point on a Circle

  • The tangent at any point of a circle is perpendicular to the radius through the point of contact.
  • Theorem 10.1: The tangent at a point P of a circle is perpendicular to the radius OP.

10.4 Summary

  1. The meaning of a tangent to a circle.
  2. The tangent to a circle is perpendicular to the radius through the point of contact.
  3. The lengths of the two tangents from an external point to a circle are equal.

Important Diagrams

Fig. 10.1: Circle and Line Interactions

  • (i) Non-intersecting line.
  • (ii) Secant intersecting at two points.
  • (iii) Tangent intersecting at one point.

Fig. 10.10: Tangents and Chords

  • Circle with center O, tangent TP, secant PQ, and radius OP.

Fig. 10.12: Circle Inscribed in a Quadrilateral

  • Points of tangency labeled R, Q, P, S.

Fig. 10.14: Triangle with Inscribed Circle

  • Triangle ABC with incircle centered at O, radius extending to point D.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

  • Misunderstanding Tangents: Students often confuse tangents with secants. Remember, a tangent touches the circle at exactly one point, while a secant intersects it at two points.
  • Drawing Tangents from Points: When asked to draw tangents from a point inside the circle, students may mistakenly think they can draw one. There are no tangents from a point inside the circle.
  • Identifying Points of Contact: Students may forget to label the points where tangents touch the circle, which is crucial for clarity in geometric proofs.

Tips for Success

  • Visualize the Concepts: Use diagrams to understand the relationships between tangents, secants, and circles. Drawing can help clarify how many tangents can be drawn from a point relative to the circle's position.
  • Practice Theorems: Familiarize yourself with key theorems, such as the tangent being perpendicular to the radius at the point of contact. This understanding is essential for solving related problems.
  • Check Your Work: When calculating lengths of tangents, ensure you apply the Pythagorean theorem correctly, especially when dealing with right triangles formed by the radius and tangent.
  • Review Common Questions: Be prepared for questions that ask about the number of tangents from various points (inside, on, and outside the circle) and the properties of tangents related to angles and lengths.

Practice & Assessment

Multiple Choice Questions

A.

They are parallel

B.

They are perpendicular

C.

They form an acute angle

D.

They form an obtuse angle
Correct Answer: B

Solution:

The tangent to a circle is perpendicular to the radius through the point of contact.

A.

The tangents are of different lengths.

B.

The tangents are equal in length.

C.

The tangents are perpendicular to each other.

D.

The tangents are parallel to each other.
Correct Answer: B

Solution:

The lengths of the two tangents from an external point to a circle are equal.

A.

7, \text{cm}

B.

8, \text{cm}

C.

9, \text{cm}

D.

10, \text{cm}
Correct Answer: C

Solution:

Let MM be the midpoint of the chord ABAB. Then AM=MB=122=6 cmAM = MB = \frac{12}{2} = 6\, \text{cm}. In the right triangle OMAOMA, OM2+AM2=OA2OM^2 + AM^2 = OA^2. Substituting the given values, 52+62=OA25^2 + 6^2 = OA^2. Solving for OAOA, we get OA=52+62=25+36=61≈7.81 cmOA = \sqrt{5^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61} \approx 7.81\, \text{cm}.

A.

10 cm

B.

11 cm

C.

9 cm

D.

12 cm
Correct Answer: B

Solution:

Let MM be the midpoint of ABAB. Then OM⊥ABOM \perp AB and AM=MB=7 cmAM = MB = 7\, \text{cm}. By the Pythagorean theorem in △OMA\triangle OMA, OA2=OM2+AM2=62+72=36+49=85OA^2 = OM^2 + AM^2 = 6^2 + 7^2 = 36 + 49 = 85. Thus, the radius OA=85≈11 cmOA = \sqrt{85} \approx 11\, \text{cm}.

A.

Secant

B.

Tangent

C.

Chord

D.

Radius
Correct Answer: B

Solution:

A tangent is a line that intersects a circle at exactly one point.

A.

Tangent

B.

Secant

C.

Non-intersecting line

D.

Chord
Correct Answer: C

Solution:

A non-intersecting line is a line that does not intersect the circle at any point.

A.

It intersects the circle at two points.

B.

It is a line that never touches the circle.

C.

It touches the circle at exactly one point.

D.

It is the longest line that can be drawn inside the circle.
Correct Answer: C

Solution:

A tangent to a circle touches the circle at exactly one point.

A.

Tangent

B.

Secant

C.

Chord

D.

Diameter
Correct Answer: C

Solution:

A line segment with both endpoints on a circle is called a chord.

A.

The lengths are equal.

B.

The lengths are different.

C.

The lengths are proportional to the radius.

D.

The lengths depend on the angle between them.
Correct Answer: A

Solution:

The lengths of the two tangents from an external point to a circle are equal.

A.

Diameter

B.

Chord

C.

Radius

D.

Tangent
Correct Answer: C

Solution:

The length from the center of a circle to any point on the circle is called the radius.

A.

Tangent

B.

Secant

C.

Chord

D.

Radius
Correct Answer: C

Solution:

A chord is a line segment with both endpoints on the circle.

A.

24 cm

B.

18 cm

C.

20 cm

D.

21 cm
Correct Answer: A

Solution:

Using the Pythagorean theorem in the right triangle OTPOTP, where OTOT is the hypotenuse, OPOP is one leg, and TPTP is the other leg, we have OT2=OP2+TP2OT^2 = OP^2 + TP^2. Therefore, TP=OT2−OP2=252−72=625−49=576=24TP = \sqrt{OT^2 - OP^2} = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24 cm.

A.

18 cm

B.

22 cm

C.

20 cm

D.

16 cm
Correct Answer: A

Solution:

In a quadrilateral with an inscribed circle, the sum of the lengths of opposite sides is equal. Thus, AB+CD=BC+DAAB + CD = BC + DA. Therefore, 8+12=10+DA⇒DA=18 cm8 + 12 = 10 + DA \Rightarrow DA = 18\, \text{cm}.

A.

The tangent is parallel to the radius.

B.

The tangent is perpendicular to the radius.

C.

The tangent bisects the radius.

D.

The tangent is equal in length to the radius.
Correct Answer: B

Solution:

The tangent to a circle is perpendicular to the radius through the point of contact.

A.

0 degrees

B.

45 degrees

C.

90 degrees

D.

180 degrees
Correct Answer: C

Solution:

The tangent to a circle is perpendicular to the radius through the point of contact, forming a 90-degree angle.

A.

11, \text{cm}

B.

12, \text{cm}

C.

13, \text{cm}

D.

14, \text{cm}
Correct Answer: B

Solution:

The length of BCBC is the sum of the segments BDBD and DCDC. Therefore, BC=BD+DC=7 cm+5 cm=12 cmBC = BD + DC = 7\, \text{cm} + 5\, \text{cm} = 12\, \text{cm}.

A.

Zero points

B.

One point

C.

Two points

D.

Three points
Correct Answer: C

Solution:

A secant of a circle intersects the circle at two points.

A.

They are unequal

B.

They are equal

C.

One is double the other

D.

They are parallel
Correct Answer: B

Solution:

The lengths of the two tangents from an external point to a circle are equal.

A.

A line that intersects the circle at exactly one point.

B.

A line that intersects the circle at exactly two points.

C.

A line that does not intersect the circle at all.

D.

A line that is perpendicular to the radius at the point of contact.
Correct Answer: B

Solution:

A secant is a line that intersects the circle at exactly two points.

A.

Tangent

B.

Secant

C.

Chord

D.

Diameter
Correct Answer: B

Solution:

A line that intersects a circle at two distinct points is called a secant.

A.

12 cm

B.

10 cm

C.

8 cm

D.

15 cm
Correct Answer: A

Solution:

Using the Pythagorean theorem in the right triangle OTPOTP, OT2=OP2+PT2OT^2 = OP^2 + PT^2. So, 132=52+PT213^2 = 5^2 + PT^2, which simplifies to PT=12 cmPT = 12 \, \text{cm}.

A.

A collection of points in a plane equidistant from a fixed point.

B.

A polygon with four sides.

C.

A line that intersects a plane at one point.

D.

A three-dimensional shape with no edges.
Correct Answer: A

Solution:

A circle is defined as the set of all points in a plane that are at a constant distance (radius) from a fixed point (center).

A.

4 cm

B.

6 cm

C.

8 cm

D.

10 cm
Correct Answer: A

Solution:

Since PQPQ is perpendicular to OPOP at PP, the midpoint of PQPQ is at PP. Thus, the distance from OO to the midpoint of PQPQ is the same as OPOP, which is 8 cm8 \, \text{cm}.

A.

It touches the circle at exactly one point.

B.

It is perpendicular to the radius at the point of contact.

C.

It can intersect the circle at two points.

D.

The lengths of tangents from an external point are equal.
Correct Answer: C

Solution:

A tangent cannot intersect the circle at two points; that is a property of a secant.

A.

Tangent

B.

Chord

C.

Secant

D.

Diameter
Correct Answer: C

Solution:

A secant is a line that intersects a circle at two distinct points.

A.

3, \text{cm}

B.

4, \text{cm}

C.

5, \text{cm}

D.

6, \text{cm}
Correct Answer: B

Solution:

Since OO is the center of the circle and ABAB is a chord, the perpendicular from OO to ABAB bisects ABAB. Let MM be the midpoint of ABAB. Then AM=MB=82=4 cmAM = MB = \frac{8}{2} = 4\, \text{cm}. In the right triangle OAMOAM, OM2+AM2=OA2OM^2 + AM^2 = OA^2. Substituting the given values, OM2+42=52OM^2 + 4^2 = 5^2. Solving for OMOM, we get OM=52−42=25−16=9=3 cmOM = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\, \text{cm}.

A.

Tangent

B.

Secant

C.

Chord

D.

Radius
Correct Answer: B

Solution:

A line that intersects a circle at two distinct points is called a secant.

A.

One

B.

Two

C.

Three

D.

Infinitely many
Correct Answer: B

Solution:

From an external point, exactly two tangents can be drawn to a circle.

A.

10 cm

B.

12 cm

C.

14 cm

D.

16 cm
Correct Answer: B

Solution:

The length of the chord PQPQ can be calculated using the Pythagorean theorem in the right triangle OPQOPQ. Since OP=6OP = 6 cm and OQ=8OQ = 8 cm, the length of PQPQ is given by PQ=OQ2−OP2=82−62=64−36=28=12PQ = \sqrt{OQ^2 - OP^2} = \sqrt{8^2 - 6^2} = \sqrt{64 - 36} = \sqrt{28} = 12 cm.

A.

12 cm

B.

10 cm

C.

8 cm

D.

15 cm
Correct Answer: A

Solution:

By the Pythagorean theorem in the right triangle OTPOTP, OT2=OP2+TP2OT^2 = OP^2 + TP^2. Substituting the given values: 132=52+TP213^2 = 5^2 + TP^2, we get 169=25+TP2169 = 25 + TP^2. Solving for TPTP, we find TP=12 cmTP = 12\, \text{cm}.

A.

13 cm

B.

12 cm

C.

5 cm

D.

17 cm
Correct Answer: A

Solution:

Since PQPQ is a tangent at PP, OPOP is perpendicular to PQPQ. By the Pythagorean theorem, OQ2=OP2+PQ2=52+122=25+144=169OQ^2 = OP^2 + PQ^2 = 5^2 + 12^2 = 25 + 144 = 169. Therefore, OQ=169=13 cmOQ = \sqrt{169} = 13\, \text{cm}.

A.

84 cm²

B.

90 cm²

C.

96 cm²

D.

100 cm²
Correct Answer: A

Solution:

The semi-perimeter ss of the triangle is 13+14+152=21 cm\frac{13 + 14 + 15}{2} = 21 \text{ cm}. Using Heron's formula, the area AA of the triangle is s(s−a)(s−b)(s−c)=21(21−13)(21−14)(21−15)=21×8×7×6=84 cm2\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = 84 \text{ cm}^2.

A.

Tangent

B.

Secant

C.

Chord

D.

Diameter
Correct Answer: B

Solution:

A line that intersects a circle at two distinct points is called a secant.

A.

10 cm

B.

12 cm

C.

15 cm

D.

20 cm
Correct Answer: B

Solution:

The perpendicular from the center OO to the chord ABAB bisects the chord. Thus, OM=8 cmOM = 8\, \text{cm} and AM=8 cmAM = 8\, \text{cm}. Using the Pythagorean theorem in △OMA\triangle OMA, OA2=OM2+AM2=82+82=64+64=128OA^2 = OM^2 + AM^2 = 8^2 + 8^2 = 64 + 64 = 128. Therefore, OA=128=82≈11.31 cmOA = \sqrt{128} = 8\sqrt{2} \approx 11.31\, \text{cm}, which rounds to 12 cm12\, \text{cm}.

A.

A line that intersects the circle at two points.

B.

A line that does not intersect the circle.

C.

A line that intersects the circle at exactly one point.

D.

A line that passes through the center of the circle.
Correct Answer: C

Solution:

A tangent to a circle is a line that intersects the circle at exactly one point.

A.

They are of different lengths.

B.

They are perpendicular to each other.

C.

They are equal in length.

D.

They pass through the center of the circle.
Correct Answer: C

Solution:

The lengths of the two tangents from an external point to a circle are equal.

A.

A line that touches the circle at one point.

B.

A line that intersects the circle at two points.

C.

A line segment with both endpoints on the circle.

D.

A line that does not intersect the circle.
Correct Answer: C

Solution:

A chord is a line segment with both endpoints on the circle.

A.

Tangent

B.

Secant

C.

Non-intersecting line

D.

Chord
Correct Answer: C

Solution:

A non-intersecting line is a line that does not intersect a circle at any point.

A.

12 cm

B.

13 cm

C.

5 cm

D.

8 cm
Correct Answer: A

Solution:

Using the Pythagorean theorem in the right triangle OTPOTP, where OPOP is the radius and OTOT is the hypotenuse, we have OT2=OP2+TP2OT^2 = OP^2 + TP^2. Substituting the known values: 132=52+TP213^2 = 5^2 + TP^2, which simplifies to 169=25+TP2169 = 25 + TP^2. Solving for TPTP, we get TP2=144TP^2 = 144, hence TP=12 cmTP = 12 \text{ cm}.

A.

6 cm

B.

7 cm

C.

8 cm

D.

9 cm
Correct Answer: B

Solution:

The length of ABAB is the sum of the segments ARAR and RBRB, which is 4 cm+3 cm=7 cm4 \, \text{cm} + 3 \, \text{cm} = 7 \, \text{cm}.

A.

Chord

B.

Diameter

C.

Radius

D.

Secant
Correct Answer: C

Solution:

A line drawn from the center of the circle to a point on the circle is called a radius.

A.

12 cm

B.

14 cm

C.

16 cm

D.

18 cm
Correct Answer: B

Solution:

Using the power of a point theorem, the length of the chord ABAB can be calculated using the formula AB=2×(r2−(OA+OB2)2)AB = \sqrt{2 \times (r^2 - (\frac{OA + OB}{2})^2)}. Substituting the given values: AB=2×(102−(7+92)2)=2×(100−64)=72=14 cmAB = \sqrt{2 \times (10^2 - (\frac{7 + 9}{2})^2)} = \sqrt{2 \times (100 - 64)} = \sqrt{72} = 14 \text{ cm}.

A.

10 cm

B.

15 cm

C.

20 cm

D.

5 cm
Correct Answer: A

Solution:

The lengths of tangents drawn from an external point to a circle are equal. Therefore, TP=TQ=10 cmTP = TQ = 10\, \text{cm}.

A.

36 cm

B.

42 cm

C.

44 cm

D.

48 cm
Correct Answer: C

Solution:

The perimeter of the quadrilateral ABCDABCD is the sum of the sides ABAB, BCBC, CDCD, and DADA. Since the circle is inscribed, AP=ASAP = AS, BQ=BPBQ = BP, CR=CQCR = CQ, and DS=DRDS = DR. Therefore, the perimeter is AB+BC+CD+DA=(AP+BP)+(BQ+CQ)+(CR+DR)+(DS+AS)=(3+4)+(4+5)+(5+6)+(6+3)=7+9+11+9=44AB + BC + CD + DA = (AP + BP) + (BQ + CQ) + (CR + DR) + (DS + AS) = (3 + 4) + (4 + 5) + (5 + 6) + (6 + 3) = 7 + 9 + 11 + 9 = 44 cm.

A.

Tangent

B.

Secant

C.

Non-intersecting line

D.

Chord
Correct Answer: C

Solution:

A line that does not intersect a circle at any point is called a non-intersecting line.

A.

14 cm

B.

12 cm

C.

16 cm

D.

10 cm
Correct Answer: A

Solution:

The length of the side BCBC is the sum of the segments BDBD and DCDC. Therefore, BC=BD+DC=8+6=14 cmBC = BD + DC = 8 + 6 = 14\, \text{cm}.

A.

Tangent

B.

Secant

C.

Chord

D.

Diameter
Correct Answer: B

Solution:

A line that intersects a circle at two distinct points is called a secant.

A.

4 cm

B.

5 cm

C.

6 cm

D.

7 cm
Correct Answer: C

Solution:

Since AA and BB are points on the circle, OAOA and OBOB are radii of the circle. Given OA=OB=5 cmOA = OB = 5\, \text{cm}, and AB=6 cmAB = 6\, \text{cm}, the length of the chord ABAB is indeed 6 cm6\, \text{cm} as given.

A.

A tangent is parallel to the radius of the circle.

B.

A tangent is perpendicular to the radius of the circle at the point of contact.

C.

A tangent is a line that passes through the center of the circle.

D.

A tangent is a line that intersects the circle at two points.
Correct Answer: B

Solution:

The tangent to a circle is perpendicular to the radius through the point of contact.

A.

12 cm

B.

9 cm

C.

6 cm

D.

15 cm
Correct Answer: A

Solution:

Using the Pythagorean theorem in the right triangle OTPOTP, where TPTP is the tangent and OTOT is the hypotenuse, we have OP=OT2−TP2=152−92=225−81=144=12OP = \sqrt{OT^2 - TP^2} = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12 cm.

A.

7 cm

B.

4 cm

C.

3 cm

D.

1 cm
Correct Answer: A

Solution:

The length of PBPB is the sum of PAPA and ABAB. Therefore, PB=PA+AB=3+4=7 cmPB = PA + AB = 3 + 4 = 7\, \text{cm}.

A.

6 cm

B.

8 cm

C.

9 cm

D.

12 cm
Correct Answer: B

Solution:

Let MM be the midpoint of the chord ABAB. Since ABAB is perpendicular to OAOA, triangle OAMOAM is a right triangle with OMOM as the hypotenuse. The length of AMAM is half of ABAB, which is 1212 cm. Using the Pythagorean theorem, OM=OA2+AM2=102+122=100+144=244=8OM = \sqrt{OA^2 + AM^2} = \sqrt{10^2 + 12^2} = \sqrt{100 + 144} = \sqrt{244} = 8 cm.

A.

0 degrees

B.

45 degrees

C.

90 degrees

D.

180 degrees
Correct Answer: C

Solution:

The tangent to a circle is perpendicular to the radius at the point of contact, making the angle 90 degrees.

A.

TP is longer than TQ

B.

TP is shorter than TQ

C.

TP is equal to TQ

D.

TP and TQ are unrelated
Correct Answer: C

Solution:

The lengths of the two tangents drawn from an external point to a circle are equal.

A.

12 cm

B.

13 cm

C.

5 cm

D.

8 cm
Correct Answer: A

Solution:

Using the Pythagorean theorem in the right triangle OTPOTP, where OPOP is the radius and OTOT is the hypotenuse, we have TP=OT2−OP2=132−52=169−25=144=12TP = \sqrt{OT^2 - OP^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 cm.

A.

Secant

B.

Chord

C.

Tangent

D.

Diameter
Correct Answer: C

Solution:

A tangent is a line that intersects a circle at exactly one point.

A.

3, \text{cm}

B.

4, \text{cm}

C.

5, \text{cm}

D.

9, \text{cm}
Correct Answer: A

Solution:

Since TPTP is a tangent to the circle at PP, OP⊥TPOP \perp TP. By the Pythagorean theorem, OT2=OP2+TP2OT^2 = OP^2 + TP^2. Substituting the given values, 52=42+TP25^2 = 4^2 + TP^2. Solving for TPTP, we get TP=52−42=25−16=9=3 cmTP = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\, \text{cm}.

A.

7, \text{cm}

B.

8, \text{cm}

C.

9, \text{cm}

D.

10, \text{cm}
Correct Answer: C

Solution:

Let MM be the midpoint of the chord ABAB. Then AM=MB=102=5 cmAM = MB = \frac{10}{2} = 5\, \text{cm}. In the right triangle OMAOMA, OM2+AM2=OA2OM^2 + AM^2 = OA^2. Substituting the given values, 62+52=OA26^2 + 5^2 = OA^2. Solving for OAOA, we get OA=62+52=36+25=61≈7.81 cmOA = \sqrt{6^2 + 5^2} = \sqrt{36 + 25} = \sqrt{61} \approx 7.81\, \text{cm}.

A.

The tangents are unequal in length.

B.

The tangents are equal in length.

C.

One tangent is twice the length of the other.

D.

The tangents are perpendicular to each other.
Correct Answer: B

Solution:

The lengths of the two tangents drawn from an external point to a circle are equal.

True or False

Correct Answer: False

Solution:

A circle can be inscribed in a quadrilateral only if the quadrilateral is tangential, meaning the sum of the lengths of opposite sides are equal.

Correct Answer: True

Solution:

An inscribed circle is tangent to each side of the quadrilateral, touching all four sides.

Correct Answer: True

Solution:

From an external point, two tangents can be drawn to a circle, and they are equal in length.

Correct Answer: False

Solution:

A line that intersects a circle at exactly one point is called a tangent, not a secant.

Correct Answer: True

Solution:

The lengths of the tangents from an external point to a circle are always equal.

Correct Answer: True

Solution:

The tangent to a circle is perpendicular to the radius through the point of contact.

Correct Answer: True

Solution:

An inscribed circle is tangent to each side of the triangle, touching each side at exactly one point.

Correct Answer: False

Solution:

There can only be one tangent at any given point on a circle.

Correct Answer: True

Solution:

The lengths of the two tangents from an external point to a circle are equal.

Correct Answer: True

Solution:

By definition, a secant is a line that intersects a circle at two distinct points.

Correct Answer: True

Solution:

A non-intersecting line does not touch the circle at any point, hence it has no common points with the circle.

Correct Answer: True

Solution:

The radius drawn to the point of contact of a tangent is always perpendicular to the tangent.

Correct Answer: True

Solution:

The tangent to a circle is always perpendicular to the radius at the point of contact.

Correct Answer: False

Solution:

A line that does not intersect a circle at any point is called a non-intersecting line, not a tangent.

Correct Answer: False

Solution:

The lengths of the two tangents drawn from an external point to a circle are equal.

Correct Answer: False

Solution:

A line that intersects a circle at one point is called a tangent, not a secant.

Correct Answer: True

Solution:

By definition, a circle is the collection of all points in a plane that are equidistant from a fixed point, known as the center.

Correct Answer: False

Solution:

A tangent to a circle intersects the circle at exactly one point, not two.

Correct Answer: True

Solution:

A secant is defined as a line that intersects a circle at exactly two points.

Correct Answer: False

Solution:

A chord is not necessarily longer than the radius. It can be shorter, equal, or longer depending on its position relative to the center.

Correct Answer: False

Solution:

A line that does not intersect a circle at any point is called a non-intersecting line, not a tangent.

Correct Answer: True

Solution:

At any given point on a circle, there can be only one tangent line.

Correct Answer: False

Solution:

There is only one tangent at a point of the circle.

Correct Answer: True

Solution:

According to the excerpt, a line that does not have any common points with a circle is termed a non-intersecting line.

Correct Answer: False

Solution:

A line that intersects a circle at exactly two points is called a secant, not a tangent.

Correct Answer: False

Solution:

A tangent to a circle intersects the circle at exactly one point.

Correct Answer: False

Solution:

A tangent to a circle touches the circle at exactly one point, not two.

Correct Answer: False

Solution:

A secant line intersects a circle at exactly two points.

Correct Answer: True

Solution:

By definition, the tangent to a circle is perpendicular to the radius at the point where the tangent touches the circle.

Correct Answer: True

Solution:

The tangent from an external point to a circle is always longer than the radius, as it extends from outside the circle to the point of tangency.

Correct Answer: False

Solution:

A line that does not intersect a circle is called a non-intersecting line, not a secant. A secant intersects the circle at two points.