TANGENTS FROM AN EXTERNAL POINT

SUBJECT :MATHEMATICS

CLASS : SS 2

TERM :THIRD TERM

WEEK ONE

TOPIC:TANGENTS FROM AN EXTERNAL POINT

Theorem:

The tangents to a circle from an external point are equal.

To understand this concept, we must first understand what a tangent is. A tangent is a line that touches a circle at only one point, and is perpendicular to the radius at that point.

Now, let us consider an external point P (as shown in the figure below) to a circle with center O and radius r. We will draw two tangents, PA and PB, from point P to the circle.

 

We have to prove that the length of both tangents, PA and PB, is equal.

To do so, let’s draw a line from the center of the circle, O, to the point P, as shown below:

 

Now, we can see that PA and PB are the two radii of the circle OPA and OPB, respectively.

Since the length of the radii of a circle is always constant, we can say that:

OA = OB = r (since both are radii of the circle)

Therefore, we can conclude that PA = PB.

Hence, we have proved that the tangents drawn from an external point to a circle are equal.

Given: a point T outside a circle, centre O, TA and TB are tangents to the circle at A and B.

To prove:  |TA| = |TB|

Construction:  Join OA, OB and OT

In  ∆s OAT and OBT

OAT = OBT = 900 (radius    tangent)

|OA| = |OB|    (radii)

|OT| =  |OT|    (common side)

∆OAT = ∆OBT (RHS)

|TA| =  |TB|

 

Note that <AOT = <BOT and <ATO = <BTO hence the line joining the external point to the centre of the circle bisects the angle between the tangents and the angle between the radii drawn to the points of contact of the tangents.

 

Worked Examples :

Example 1: A circle has a radius of 5 cm. A tangent is drawn from a point P outside the circle to the circle, and it touches the circle at point Q. Find the length of PQ.

Solution: Let O be the center of the circle. We draw a line from O to P, and extend it to meet the tangent line at R.

 

Now, we can see that the line segment OR is the radius of the circle, and OP is the hypotenuse of right triangle OPR. We can use the Pythagorean theorem to find OP:

OP² = OR² + PR² OP² = 5² + 12² (since OR = 5 and PR = 12) OP² = 169 OP = √169 = 13

Therefore, PQ = OP – OQ = 13 – 5 = 8 cm.

Hence, the length of PQ is 8 cm.

Example 2: A circle has a radius of 6 cm. A point P is located outside the circle such that the length of the tangent from P to the circle is 10 cm. Find the distance between the point P and the center of the circle.

Solution: Let O be the center of the circle, and let Q be the point where the tangent line meets the circle.

 

We can see that OQ is perpendicular to the tangent line, and it is also the radius of the circle. Therefore, OQ = 6 cm.

We can use the Pythagorean theorem to find OP:

OP² = OQ² + PQ² OP² = 6² + 10² (since PQ = 10) OP² = 136 OP = √136 = 2√34

Therefore, the distance between point P and the center of the circle is 2√34 – 6 cm.

Hence, the distance is approximately 4.63 cm.

Example 3: A circle has a radius of 4 cm. A point P is located outside the circle such that the length of the tangent from P to the circle is 8 cm. Find the length of the line segment joining the center of the circle to the point P.

Solution: Let O be the center of the circle, and let Q be the point where the tangent line meets the circle

Example-3

We can see that OQ is perpendicular to the tangent line, and it is also the radius of the circle. Therefore, OQ = 4 cm.

We can use the Pythagorean theorem to find OP:

OP² = OQ² + PQ² OP² = 4² + 8² (since PQ = 8) OP² = 80 OP = √80 = 4√5

Therefore, the length of the line segment joining the center of the circle to the point P is 4√5 cm.

Hence, the length is approximately 8.94 cm.

Example 4.PQR are three points on a circle Centre O. The tangent at P and Q meet at T. If   < PTQ = 620 calculate PRQ.

 

Solution

Join OP and OQ

In quadrilateral TQOP

<OQT = <OPT = 900 (radius 1 tangent)

POQ = 3600 – (900 + 900 + 620) sum of angle in a quadrilateral)

POQ = 3600 – 2420

POQ = 1180

PRQ = 1180 =  590 (2x angle at circumference = angle at centre)

      2

PR1QR is a cyclic quadrilateral

R + R1 = 1800 (opp. angles of  a cyclic quadrilateral )

R1 = 1800 – R

R1 = 1800 – 590

R1 = 1210

 PRQ  =  590  or 1210

 

Evaluation 

1. ABC are three points on a circle, centre O such that <BAC = 370, the tangents at B and C meet at T. Calculate  < BTC. 

 

GENERAL EVALUATION/REVISION QUESTIONS

1. What is a tangent to a circle? a) A line that intersects the circle at two points. b) A line that intersects the circle at one point. c) A line that does not intersect the circle.
2. If a tangent is drawn from an external point to a circle, how many points of intersection are there? a) 0 b) 1 c) 2
3. If a tangent is drawn from an external point to a circle, what is the relationship between the tangent and the radius at the point of intersection? a) They are parallel. b) They are perpendicular. c) They are at an angle.
4. What is the name of the point where the tangent intersects the circle? a) Center b) Vertex c) Point of contact
5. If two tangents are drawn from an external point to a circle, what is the relationship between the two tangents? a) They are perpendicular. b) They are parallel. c) They are equal in length.
6. If the length of a tangent from a point outside a circle to the circle is 5 cm, and the distance between the point and the center of the circle is 8 cm, what is the radius of the circle? a) 13 cm b) 12 cm c) 10 cm
7. If the length of a tangent from a point outside a circle to the circle is 15 cm, and the radius of the circle is 6 cm, what is the distance between the point and the center of the circle? a) 9 cm b) 12 cm c) 18 cm
8. If the length of a tangent from a point outside a circle to the circle is 10 cm, and the radius of the circle is 8 cm, what is the length of the chord joining the points of contact of the tangents from the point? a) 16 cm b) 20 cm c) 24 cm
9. If a tangent is drawn from an external point to a circle, and the length of the tangent is 10 cm, what is the length of the line segment joining the center of the circle to the point? a) 5 cm b) 6 cm c) 8 cm
10. If two tangents are drawn from a point outside a circle, and the length of one tangent is 6 cm, what is the length of the other tangent? a) 6 cm b) 8 cm c) 10 cm

1. AB is a chord and O is the centre of a circle. If AOB = 780 calculate the obtuse angle between AB and the tangent B.
2. The dimension of a cuboid metal is 24cm by 21cm by 10cm, if the cuboid is melted and used in making a cylinder whose base radius is 15cm find the height of the cylinder.
3. The volume of a cylinder is 3600cm3 and its radius is 10cm calculate its

(a) curve surface area

(b) total surface area

 

READING ASSIGNMENT

Essential Mathematics, pages149-151, numbers 11-20.

WEEKEND ASSIGNMENT

Use the diagram below to answer the questions. 

 

1.If  < ATO  =  360 ,calculate < ABO.

(a) 360     (b) 720      (c) 180     (d) 440

2.If <ABT  = 570, calculate   < AOT   (a) 1140     (b) 570    (c) 330     (d) 1230

3.If< BTO = 440, calculate <TAX   (a) 880   (b) 440    (c) 460   (d) 920

4.If   |AB| =  18cm and  |TB| = 15cm, calculate |TX|

    (a) 180     (b) 330     (c) 780     (d) 120

5.If < AOT = 470, calculate ABO   (a) 470     (b) 940      (c) 1330       (d) 430

 

THEORY

1.O is the centre of a circle and two tangents from a point T touch the centre at A and B. BT is produced to C. If <AOT = 670.calculate < ATC.

First, we know that the tangent to a circle at any point is perpendicular to the radius drawn to that point. Therefore, OB and OC are both perpendicular to BT and CT respectively, and so we have the following diagram:

Since O is the center of the circle, OB and OC are both radii of the circle, and so they are equal.

Also, since angle BAC = 370 degrees, we know that angle BOC (the angle subtended by the arc BC) is twice that, or 740 degrees.

Now, we can use the fact that the sum of the angles in a triangle is 180 degrees to find angle BTC.

In triangle BTC, we have:

angle TBC + angle TCB + angle BTC = 180 degrees

Since OB and OC are radii of the circle, angle TBC = angle TCB, and we can let x be the value of each of these angles.

Thus, we have:

2x + angle BTC = 180 degrees

And from above, we know that angle BOC = 740 degrees, so:

angle BTC = 180 degrees – 2x = 180 degrees – (1/2)(740 degrees) = 100 degrees

Therefore, <BTC = 100 degrees

2.AD is a diameter of a circle,AB is a chord and AT is a tangent. a) State  the size of  <ADBb)If BAT is an acute angle of  x0,find  the  size  of DAB in terms of   x.

 

Given: O is the center of a circle, TA and TB are tangents from point T, and BT is produced to C. To find: the measure of angle ATC.

Construction: Join OC and TO.

Proof: Since TA and TB are tangents to the circle, we know that OA and OB are radii of the circle and are perpendicular to TA and TB, respectively. So, we have: ∠OAT = ∠OBT = 90° (as radii are perpendicular to the tangent)

Also, we have: ∠OAB = 1/2 ∠OTB (angle at the center is twice the angle at the circumference)

And, since ATCB is a cyclic quadrilateral (as opposite angles sum up to 180 degrees), ∠ATC + ∠ABC = 180°

Now, we can use these facts to find the measure of angle ATC. ∠AOT = ∠OAT + ∠OTA = ∠OBT + ∠OTB = ∠BOT

So, ∠AOT = ∠BOT = 90° Now, in triangle OTC, ∠OTC = 180° – ∠OCT – ∠OTB = 180° – 90° – 90° = 0°

Hence, the line segment TC is parallel to the tangent AB. Therefore, ∠ABC = ∠ATC, as they are alternate angles between parallel lines. So, ∠ATC + ∠ATB = 180° (from the cyclic quadrilateral) ∠ATC + 90° = 180° (∠ATB = 90°, as it is a tangent to the circle) Therefore, ∠ATC = 90°.

Now, in triangle OTC, ∠TOC = 180° – ∠OTC – ∠OTC = 180° – 0° – 0° = 180°

Therefore, ∠ATC + ∠TOC = 270° But, ∠AOT = 67°, so ∠TOC = ∠AOT = 67° Hence, ∠ATC = 90° – ∠TOC = 90° – 67° = 23°

Therefore, the measure of angle ATC is 23 degrees.

Given: AD is a diameter of a circle, AB is a chord, and AT is a tangent. To find: (a) the measure of angle ADB, (b) the measure of angle DAB in terms of x.

Construction: Join BD.

Proof: (a) In circle O, ∠OAD = 90° (as AD is a diameter) So, ∠OAB = 1/2 ∠ADB (angle at the center is twice the angle at the circumference)

Also, ∠BAT = 90° (as AT is a tangent) Therefore, ∠BAD = 90° – ∠BAT = 90° – x0

In right triangle ADB, ∠ADB + ∠BAD + ∠A = 180° (sum of angles in a triangle) So, ∠ADB = 180° – ∠BAD – ∠A = 180° – (90° – x0) – (1/2 ∠ADB) Simplifying, we get: 3/2 ∠ADB = 90° + x0

 

Evaluation

1. In the given figure, what is the value of <AOT? (a) 90° (b) 120° (c) 130° (d) 170°
2. In the given figure, what is the value of <ATC? (a) 30° (b) 40° (c) 50° (d) 60°
3. In the given figure, what is the value of <ACB? (a) 30° (b) 45° (c) 60° (d) 90°
4. In the given figure, what is the value of <ADB? (a) 45° (b) 60° (c) 90° (d) 180°
5. In the given figure, what is the value of <ABD? (a) 30° (b) 45° (c) 60° (d) 90°
6. In the given figure, what is the value of <BAT? (a) x0 (b) 90° – x0 (c) 180° – x0 (d) 360° – x0
7. In the given figure, what is the value of <DAB? (a) 30° – 2/3 x0 (b) 45° – 2/3 x0 (c) 60° – 2/3 x0 (d) 90° – 2/3 x0
8. In the given figure, what is the value of <ADB + <ACB? (a) 90° (b) 120° (c) 150° (d) 180°
9. In the given figure, what is the value of <ABD + <BAD? (a) 90° (b) 120° (c) 150° (d) 180°
10. In the given figure, what is the relationship between |TA| and |TB|? (a) |TA| = |TB| (b) |TA| > |TB| (c) |TA| < |TB| (d) Cannot be determined

Lesson Plan Presentation:

The Tangents to a Circle from an External Point are Equal

Objectives:

• To understand the concept of tangents to a circle from an external point
• To learn how to prove that the tangents to a circle from an external point are equal
• To solve problems related to this concept

Materials:

• Whiteboard and markers
• Printed handouts of examples and practice questions
• Calculators
• Compasses

Procedure:

I. Introduction

• Greet the class and introduce the topic: The Tangents to a Circle from an External Point are Equal
• State the objectives of the lesson

II. Definition and Explanation

• Define what tangents are and how they relate to circles
• Explain what is meant by the tangents to a circle from an external point
• Show the class the diagram of a circle with tangents drawn from an external point

III. Theorem and Proof

• State the theorem: The tangents to a circle from an external point are equal
• Explain how to prove this theorem using construction and congruency
• Provide a step-by-step example of a proof using a diagram on the whiteboard

IV. Examples

• Present several examples of problems related to this theorem
• Demonstrate how to solve each problem using the theorem and the proof method taught
• Provide printed handouts for the students to work on their own and discuss in groups

V. Practice Questions

• Give the class a set of practice questions to solve individually
• Provide assistance and guidance as needed
• Review the answers and solutions as a class

VI. Conclusion

• Summarize the key points of the lesson
• Answer any remaining questions from the class
• Assign homework if necessary

Assessment:

• Monitor student participation and understanding during the lesson
• Evaluate the student’s ability to solve practice problems and homework assignments
• Provide feedback on their progress and areas for improvement

Closing:

• Thank the class for their attention and participation
• Encourage them to ask questions or seek help if they need further clarification
• Remind them of the homework assignments and when they are due