Circle Properties (Part 2) | Tangents & Four Centres

Synopsis

This article covers advanced circle properties including tangents and the four centres of a triangle – a high-frequency topic in DSE Paper 1 Section A(2) and Section B worth 4–7 marks. You will learn the tangent theorems (tangent perpendicular to radius, tangent lengths from an external point, and the alternate segment theorem), and the definitions and properties of the circumcentre, incentre, centroid, and orthocentre of a triangle, with applications in circle geometry. The article includes step-by-step worked examples, DSE exam techniques, practice questions. These are high-level skills that distinguish top-performing students.


Learning Objectives

By the end of this article, you should be able to:

  • Apply the tangent-radius theorem (tangent ⊥ radius at point of tangency).
  • Apply the tangent length theorem (equal tangents from an external point).
  • Apply the alternate segment theorem (angle between tangent and chord equals angle in alternate segment).
  • Define and identify the circumcentre (circumcircle centre), incentre (incircle centre), centroid, and orthocentre of a triangle.
  • Apply these concepts to solve DSE-style geometry problems involving circles and triangles.
  • Prove geometric relationships using tangent and centre properties.

1. Introduction:

Tangents and the four centres of a triangle appear every year in DSE Paper 1 Section A(2) and Section B. Questions may ask you to:

  • Find unknown angles or lengths using tangent theorems
  • Identify the circumcentre or incentre from given conditions
  • Apply the alternate segment theorem in proofs
  • Solve multi-step problems involving tangents and centres

These are high-level skills that can help you secure top marks.

DSE Exam Tip

Tangent questions often appear in Section A(2) as short-answer questions worth 3–4 marks, and in Section B as longer proof questions worth 5–7 marks. The four centres are often tested in combination with circle theorems.

2. Tangent Theorems

Theorem 1: Tangent-Radius Theorem

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

If \( OT \) is a radius and \( PT \) is a tangent at \( T \), then \( OT \perp PT \).

Theorem 2: Tangent Lengths from an External Point

From a point outside a circle, the two tangent lengths are equal.

If \( PA \) and \( PB \) are tangents from \( P \) to the circle, then \( PA = PB \).

Theorem 3: Alternate Segment Theorem

The angle between a tangent and a chord is equal to the angle in the alternate segment (the angle subtended by the chord at the circumference on the opposite side).

If \( PT \) is a tangent at \( T \) and \( TA \) is a chord, then \( \angle PTA = \angle TBA \) where \( B \) is any point on the circle in the alternate segment.

DSE Memory Aid

Alternate segment theorem: The angle between the tangent and the chord equals the angle in the opposite arc (the "alternate" segment).

3. Worked Examples on Tangents

Example 1: Tangent-Radius

Question: In the figure, \( PT \) is tangent to the circle at \( T \), and \( OT = 5 \) cm, \( OP = 13 \) cm. Find the length of \( PT \).

Solution

By tangent-radius theorem, \( OT \perp PT \), so triangle \( OTP \) is right-angled.

Using Pythagoras: \( OP^2 = OT^2 + PT^2 \)\( 13^2 = 5^2 + PT^2 \)\( PT^2 = 169 - 25 = 144 \)\( PT = 12 \) cm.

Answer: \( 12 \) cm.

Example 2: Alternate Segment Theorem

Question: In the figure, \( PA \) is tangent at \( A \). \( \angle PAB = 40^\circ \). Find \( \angle ACB \).

Solution

By the alternate segment theorem, \( \angle PAB = \angle ACB \) (angle in the alternate segment).

Thus \( \angle ACB = 40^\circ \).

Answer: \( 40^\circ \).

4. The Four Centres of a Triangle

A triangle has four important centres. Each has a distinct definition and geometric significance:

CentreDefinitionCircle Associated
Circumcentre Intersection of perpendicular bisectors of the sides Circumcircle (passes through all vertices)
Incentre Intersection of angle bisectors Incircle (tangent to all sides)
Centroid Intersection of medians
Orthocentre Intersection of altitudes (perpendicular from vertex to opposite side)
Common DSE Trap

Don't confuse the circumcentre (equidistant from vertices) with the incentre (equidistant from sides). The circumcentre is the centre of the circumcircle, while the incentre is the centre of the incircle.

Key Properties for DSE

  • Circumcentre: Equidistant from all three vertices. Lies on perpendicular bisectors. For an acute triangle, inside; right triangle, on the hypotenuse; obtuse, outside.
  • Incentre: Equidistant from all three sides. Always inside the triangle. The angle bisectors meet at the incentre.
  • Centroid: Divides each median in the ratio \( 2:1 \) (from vertex to midpoint). Always inside.
  • Orthocentre: Intersection of altitudes. Inside for acute, on the vertex for right, outside for obtuse.

5. Worked Examples on Four Centres

Example 1: Circumcentre

Question: Triangle \( ABC \) has vertices \( A(2,3) \), \( B(4,7) \), \( C(6,3) \). Find the circumcentre.

Solution

The circumcentre is the intersection of perpendicular bisectors. Since \( A \) and \( C \) have the same y-coordinate, the perpendicular bisector of \( AC \) is the vertical line \( x = \frac{2+6}{2}=4 \).

Midpoint of \( AB \) is \( (3,5) \), slope of \( AB \) is \( \frac{7-3}{4-2} = 2 \), so perpendicular slope = \( -\frac{1}{2} \). Equation: \( y-5 = -\frac{1}{2}(x-3) \). At \( x=4 \), \( y-5 = -\frac{1}{2}(1) = -0.5 \)\( y=4.5 \). So circumcentre is \( (4, 4.5) \).

Answer: \( (4, 4.5) \).

Example 2: Incentre

Question: In triangle \( ABC \), \( \angle A = 60^\circ \), \( \angle B = 70^\circ \). Find \( \angle AIC \) where \( I \) is the incentre.

Solution

The incentre is the intersection of angle bisectors. In triangle \( AIC \), \( \angle IAC = \frac{1}{2}\angle A = 30^\circ \), \( \angle ICA = \frac{1}{2}\angle C \). First, \( \angle C = 180^\circ - 60^\circ - 70^\circ = 50^\circ \), so \( \angle ICA = 25^\circ \).

Then \( \angle AIC = 180^\circ - 30^\circ - 25^\circ = 125^\circ \).

Answer: \( 125^\circ \).

Example 3: Centroid

Question: Triangle \( ABC \) has vertices \( (1,2) \), \( (3,6) \), \( (5,4) \). Find the centroid.

Solution

Centroid is the average of the coordinates: \( \left( \frac{1+3+5}{3}, \frac{2+6+4}{3} \right) = \left( \frac{9}{3}, \frac{12}{3} \right) = (3,4) \).

Answer: \( (3,4) \).

6. Combining Tangents and Centres

In DSE questions, tangents and centres often appear together. For example, the incentre is the centre of the incircle, which is tangent to all sides. The circumcentre is the centre of the circumcircle, which passes through all vertices.

Worked Example

Question: In triangle \( ABC \), the incircle touches \( AB \) at \( D \). If \( AD = 4 \), \( DB = 6 \), and \( AC = 8 \), find the length of \( BC \).

Solution

Using tangent lengths from the same external point: from \( A \), tangents to the incircle are equal, so \( AD = AE = 4 \) (where \( E \) is touchpoint on \( AC \)). Since \( AC = 8 \), \( EC = 8 - 4 = 4 \).

From \( B \), tangents equal: \( BD = BF = 6 \) (where \( F \) is touchpoint on \( BC \)).

From \( C \), tangents equal: \( CE = CF = 4 \).

Thus \( BC = BF + FC = 6 + 4 = 10 \).

Answer: \( 10 \).

7. DSE-Style Practice Questions

Section A(2) & Section B Style

Question 1 MC
In the figure, \( PA \) and \( PB \) are tangents to the circle at \( A \) and \( B \). If \( \angle APB = 70^\circ \), find \( \angle AOB \) where \( O \) is the centre.
A. \( 70^\circ \)     B. \( 110^\circ \)     C. \( 140^\circ \)     D. \( 55^\circ \)

Question 2 MC
Which centre of a triangle is equidistant from the three sides?
A. Circumcentre     B. Incentre     C. Centroid     D. Orthocentre

Question 3 Short Answer
In the figure, \( PT \) is tangent at \( T \). \( \angle PTQ = 50^\circ \). Find \( \angle PRQ \).

Question 4 Short Answer
Triangle \( ABC \) has vertices \( (0,0) \), \( (4,0) \), \( (0,3) \). Find the circumcentre.

Question 5 MC
The incircle of triangle \( ABC \) touches \( BC \) at \( D \). If \( BD = 3 \), \( DC = 5 \), \( AB = 7 \), find \( AC \).
A. \( 9 \)     B. \( 8 \)     C. \( 7 \)     D. \( 6 \)

8. Solutions with Explanations

Question 1: B. \( 110^\circ \)
\( \angle AOB = 180^\circ - \angle APB \) (since tangents are perpendicular to radii, quadrilateral \( AOBP \) has \( \angle OAP = \angle OBP = 90^\circ \), so \( \angle AOB + \angle APB = 180^\circ \)). Thus \( \angle AOB = 110^\circ \).
Question 2: B. Incentre
The incentre is equidistant from all three sides (distance to each side is the inradius).
Question 3: \( 50^\circ \)
By the alternate segment theorem, \( \angle PTQ = \angle PRQ \), so \( \angle PRQ = 50^\circ \).
Question 4: \( (2, 1.5) \)
The triangle is right-angled at \( (0,0) \). The circumcentre of a right triangle is the midpoint of the hypotenuse. Hypotenuse endpoints \( (4,0) \) and \( (0,3) \) → midpoint \( (2, 1.5) \).
Question 5: A. \( 9 \)
Tangent lengths: from \( A \), tangents to incircle are equal, so \( AB = AC \)? Actually, let \( x \) be tangent length from \( A \). Then \( AB = x + 3 \) (since \( BD=3 \)), \( AC = x + 5 \) (since \( DC=5 \)). Given \( AB = 7 \), \( x + 3 = 7 \)\( x=4 \). Then \( AC = 4 + 5 = 9 \).

9. Exercise

Click the following link to have
 An Exercise on Circle Properties (Part 2) | Tangents & Four Centres

Key Takeaways

What You Should Remember
  • Tangent-radius: tangent ⊥ radius at point of tangency.
  • Tangent lengths: from an external point, tangents are equal.
  • Alternate segment theorem: angle between tangent and chord equals angle in alternate segment.
  • Circumcentre: centre of circumcircle, equidistant from vertices.
  • Incentre: centre of incircle, equidistant from sides, intersection of angle bisectors.
  • Centroid: intersection of medians, divides each median 2:1.
  • Orthocentre: intersection of altitudes.
  • This topic guarantees 4–7 marks in DSE Paper 1 Section A(2) and Section B – master these skills!

Summary Checklist for Revision

  • Tangent-radius theorem
  • Equal tangents from external point
  • Alternate segment theorem
  • Circumcentre: perpendicular bisectors
  • Incentre: angle bisectors
  • Centroid: medians 2:1
  • Orthocentre: altitudes
  • Apply tangent+centre problems