Circles • Topic 3 of 3

Tangents from an External Point & Geometrical Applications

What is the Length of a Tangent Theorem? When you choose a point completely outside a circle, you can draw exactly two tangents to it. The Lengths of Tangents Theorem states: The lengths of these two tangent segments drawn from an external point to a circle are perfectly equal to each other.

To prove this key property, imagine drawing lines from the external point to the center of the circle, along with the two radii to the points of contact. This creates two distinct right-angled triangles sharing the same hypotenuse. Because both triangles have a right angle, identical radii lengths, and share a common side, they are completely congruent by the RHS (Right angle-Hypotenuse-Side) rule.

As a direct consequence of this triangle congruence:

  • The two outer tangent segments are identical in length.
  • The line joining the external point to the center perfectly bisects the angle between the two tangents.
  • The line joining the external point to the center perfectly bisects the angle between the two radii.

Quadrilateral and Angle Properties When you trace the two radii and the two external tangents together, they form a closed four-sided polygon (a quadrilateral). Since two corners of this quadrilateral are right angles (90 degrees each), the remaining two opposite angles (the angle between the tangents and the angle between the radii) must add up to exactly 180 degrees. This means they are supplementary angles.

---

DIAGRAM 1: CONGRUENT TRIANGLES FORMED BY EXTERNAL TANGENTS

                     T1 (Point of Contact 1)
                     /\
                    /  \ Radius (r)
                   /    \
                  /      \
      External   /________\ O (Center)
       Point P   \        /
                  \      /
                   \    / Radius (r)
                    \  /
                     \/
                     T2 (Point of Contact 2)
                     
    Triangle PT1O is perfectly congruent to Triangle PT2O. Therefore, PT1 = PT2.

DIAGRAM 2: SUPPLEMENTARY ANGLES IN TANGENT QUADRILATERALS

                     T1
                    X\
                   /  \ 90°
                  /    \
           Angle /______\ Center O (Angle y)
             (x) \      /
                  \    / 90°
                   \  /
                    X/
                     T2
                     
    Angle (x) + Angle (y) + 90° + 90° = 360°  =>  Angle (x) + Angle (y) = 180°

DIAGRAM 3: RULER AND COMPASS CONSTRUCTION CONCEPT
                    
                     T1
                    / \
                   /   \  [Draw Circle with diameter OP]
                  /  .  \
                 P .  O  .
                  \  .  /
                   \   /
                    \ /
                     T2
Solved Examples
7
Worked Example
From a point P located outside a circle with center O, two tangents PA and PB are drawn to touch the circle at points A and B respectively. If the length of tangent segment PA is measured to be 14 cm, what is the length of the tangent segment PB?
Solution
  1. Step 1: Recall the fundamental theorem regarding external tangents.*
  2. The theorem states that the lengths of two separate tangents drawn from the exact same external point to a single circle must be equal.*
  3. Step 2: Match the segments to the theorem statement.*
  4. The segments PA and PB are both tangents drawn from the identical external point P.*
  5. Therefore, length of PB must equal the length of PA.*
  6. Step 3: State the final dimension.*
  7. Since PA = 14 cm, it must be true that PB = 14 cm.*
  8. Answer: The length of PB is 14 cm.
8
Worked Example
Two tangents PA and PB are drawn to a circle centered at O from an external point P. If the angle between the two tangents (Angle APB) is measured to be 70 degrees, calculate the measure of the central angle formed between the two radii (Angle AOB).
Solution
  1. Step 1: Identify the geometric structure and angle relationships.*
  2. The lines OA, AP, PB, and BO form a quadrilateral OAPB.*
  3. The angles OAP and OBP are both 90-degree right angles because radii are perpendicular to tangents.*
  4. Step 2: Apply the supplementary angle rule for external tangent configurations.*
  5. The sum of the central angle (Angle AOB) and the external angle between the tangents (Angle APB) is always exactly 180 degrees.*
  6. Angle AOB + Angle APB = 180 degrees*
  7. Step 3: Substitute the known value and solve for the unknown angle.*
  8. Angle AOB + 70 degrees = 180 degrees*
  9. Angle AOB = 180 degrees - 70 degrees = 110 degrees.*
  10. Answer: Angle AOB is 110 degrees.
9
Worked Example
A circle is inscribed perfectly inside a triangle ABC, touching the sides AB, BC, and CA at points D, E, and F respectively. If the lengths of the triangle sides are measured such that AD = 5 cm, BE = 6 cm, and CF = 7 cm, calculate the complete perimeter of the triangle ABC.
Solution
  1. Step 1: Use the external tangents theorem at each individual vertex of the triangle.*
  2. Vertex A acts as an external point yielding two equal tangents: AD = AF = 5 cm.*
  3. Vertex B acts as an external point yielding two equal tangents: BD = BE = 6 cm.*
  4. Vertex C acts as an external point yielding two equal tangents: CE = CF = 7 cm.*
  5. Step 2: Determine the complete length of each of the three main sides of the triangle.*
  6. Side AB = AD + BD = 5 cm + 6 cm = 11 cm*
  7. Side BC = BE + CE = 6 cm + 7 cm = 13 cm*
  8. Side CA = AF + CF = 5 cm + 7 cm = 12 cm*
  9. Step 3: Sum the side lengths to calculate the total perimeter.*
  10. Perimeter = Side AB + Side BC + Side CA*
  11. Perimeter = 11 cm + 13 cm + 12 cm = 36 cm.*
  12. Answer: The perimeter of triangle ABC is 36 cm.
  13. --

Key Points

  • Tangent segments drawn onto a single circle from the same external point share identical lengths.
  • The line connecting the external point to the center acts as a perfect angle bisector for the system.
  • The angle between the two tangents and the angle between the corresponding radii are supplementary (sum to 180 degrees).
  • Triangles inscribed or circumscribed around circles can be solved easily by splitting sides into pairs of equal tangent segments.
  • To construct these tangents using a compass, you find the midpoint of the line to the center and draw an auxiliary circle.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The two tangents drawn from an external point are:
Explanation: Equal in length.
Q2.The length of a tangent from a point at distance $d$, radius $r$, is:
Explanation: From the right triangle.
Q3.If a tangent has length $8$ and radius $6$, the distance of the point from the centre is:
Explanation: $\sqrt{8^2+6^2}=10$.
Q4.An incircle of a triangle touches each side at:
Explanation: One point per side.
Practice more MCQs → 📝 Assignment · 35 marks