CIE IGCSE Mathematics (0580) Circle theorems I Study Notes - New Syllabus
CIE IGCSE Mathematics (0580) Circle theorems Study Notes
LEARNING OBJECTIVE
- Understanding Key Circle Theorems
Key Concepts:
- Circle Theorems
Circle Theorems
Circle Theorems
Circle theorems help in solving geometric problems involving angles, chords, tangents, and arcs.
Key Theorems :
- Angle in a Semicircle: Any angle formed in a semicircle is a right angle (90°).
- Tangent-Radius Angle: The angle between a tangent and the radius drawn to the point of contact is 90°.
1. Angle in a Semicircle
If a triangle is drawn such that one side is the diameter of the circle and the third point lies on the circle, then the angle opposite the diameter is always a right angle.
Theorem 1: Angle in a Semicircle = 90°
Any triangle inscribed in a semicircle with the diameter as one side forms a right angle at the opposite vertex.
Proof:
- Let the circle have centre \( O \), and let \( AB \) be the diameter of the circle.
- Let point \( C \) lie on the circle, forming triangle \( ABC \).
- Join points \( O \) to \( A \), \( B \), and \( C \), so \( OA = OB = OC \) (radii of the circle).
- Then triangles \( OAC \) and \( OBC \) are isosceles.
- This gives:
\( \angle OAC = \angle OCA \) and \( \angle OBC = \angle OCB \) - Since \( \angle AOB = 180^\circ \) (straight angle on diameter), and total angle in triangle is 180°, we have:
\( \angle ACB = 180^\circ – (\angle CAB + \angle ABC) = 90^\circ \)
Conclusion: \( \angle ACB = 90^\circ \)
Example:
In a circle with diameter AB, point C lies on the circumference. What is angle ACB?
▶️ Answer/Explanation
Since AB is the diameter, and C lies on the circle, triangle ACB is formed in a semicircle.
Therefore, by the theorem: \( \angle ACB = 90^\circ \)
Reason: Angle in a semicircle is a right angle.
Example:
O is the centre of the circle. Find the value of x.
▶️Answer/Explanation
\( \angle ABC = 90^\circ \quad \text{(angle in a semicircle = }90^\circ\text{)} \)
\( 63^\circ + 90^\circ + x = 180^\circ \quad \text{(sum of angles in a triangle)} \)
\( x = 27^\circ \)
2. Angle between Tangent and Radius
When a tangent touches a circle at a point, the angle it makes with the radius drawn to that point is always 90°. Tangents are perpendicular to the radii at the point of contact.
Theorem 2: Tangent and Radius Meet at 90°
The angle between a radius and a tangent to the circle at the point of contact is always 90°.
Proof:
- Let the circle have centre \( O \), and let the tangent touch the circle at point \( P \).
- Draw radius \( OP \).
- Suppose \( OP \) is not perpendicular to the tangent.
- Then there exists another line from \( O \) to the tangent which is shorter than \( OP \).
- This contradicts the definition of a circle: all radii are equal, and the radius is the shortest distance from the centre to the circle.
- Hence, only when the radius is perpendicular does it touch the tangent at exactly one point.
Conclusion: \( \angle OPB = 90^\circ \)
Example:
A tangent touches a circle at point P. O is the centre of the circle. What is the angle between radius OP and the tangent?
▶️ Answer/Explanation
The radius OP meets the tangent at point P.
Therefore, by the theorem: \( \angle \text{between radius and tangent} = 90^\circ \)
Reason: Tangent is perpendicular to the radius at the point of contact.
Example:
Two concentric circles are of radii 13 cm and 5 cm. Find the length of the chord of the larger circle which touches the smaller circle.
In the figure, \( O \) is the common center of the given concentric circles.
▶️Answer/Explanation
\( AB \) is the chord of the larger circle such that it is tangent to the smaller circle at point \( P \).
Since \( OP \) is the radius of the smaller circle,
\( OP \perp AB \Rightarrow \angle APO = 90^\circ \)
Also, radius perpendicular to the chord bisects the chord.
So, \( OP \) bisects \( AB \), hence:
\( AP = \frac{1}{2} AB \)
Now, in right-angled triangle \( \triangle APO \):
\( AP^2 = OA^2 – OP^2 \)
\( = (13)^2 – (5)^2 = 169 – 25 = 144 \)
\( \Rightarrow AP = \sqrt{144} = 12 \, \text{cm} \)
Using \( AP = \frac{1}{2} AB \):
\( 12 = \frac{1}{2} AB \Rightarrow AB = 2 \times 12 = \boxed{24 \, \text{cm}} \)
Example:
A line \( CB \) is tangent to the circle at point \( C \). The radius \( AC \) is perpendicular to the tangent. If \( \angle CAB = 60^\circ \), find the value of angle \( \theta \).
▶️Answer/Explanation
Given that \( AC \) is perpendicular to the tangent at point \( C \), we know:
\( \angle ACB = 90^\circ \)
Using triangle angle sum property in \( \triangle ABC \):
\( \angle CAB + \angle ACB + \theta = 180^\circ \)
\( 60^\circ + 90^\circ + \theta = 180^\circ \)
\( \theta = 180^\circ – 150^\circ = \boxed{30^\circ} \)
More Circle Theorems
3. Angle at the Centre is Twice the Angle at the Circumference
The angle subtended by an arc at the centre of a circle is twice the angle subtended at any point on the remaining part of the circumference.
Theorem : \( \angle AOB = 2 \angle ACB \)
If points A, B, and C lie on a circle and O is the centre, then the angle at the centre \( \angle AOB \) is twice the angle at the circumference \( \angle ACB \).
Proof:
- Let O be the centre of the circle, and let A, B, and C be points on the circle such that angle \( \angle ACB \) is subtended by arc AB.
- Join \( OA \), \( OB \), and \( OC \). This divides triangle AOB into two isosceles triangles: \( \triangle OAC \) and \( \triangle OBC \).
- Let \( \angle ACO = \angle CAO = x \), and \( \angle BCO = \angle CBO = y \).
- Then the angle at centre is \( \angle AOB = 2x + 2y = 2(x + y) \), and angle at circumference is \( \angle ACB = x + y \).
Conclusion: \( \angle AOB = 2 \angle ACB \)
Example:
In the circle below, Find x.
▶️ Answer/Explanation
By the theorem: Angle at the centre is twice the angle at the circumference.
\( \angle BAC = 2 \angle BOC \Rightarrow \angle x = \frac{106^\circ}{2} = \boxed{53^\circ} \)
Example:
In a circle, the angle subtended by an arc at the centre is 114°. Find the angle subtended by the same arc at the circumference.
▶️Answer/Explanation
By the theorem:
\( \angle \text{at centre} = 2 \times \angle \text{at circumference} \)
So, \( \angle \text{at circumference} = \frac{114^\circ}{2} = \boxed{57^\circ} \)
4. Angles in the Same Segment are Equal
Angles formed in the same segment of a circle and subtended by the same chord are always equal, regardless of their positions on the arc.
Theorem : \( \angle APB = \angle AQB \)
If two angles are formed on the same side of a chord (i.e., in the same segment), and they subtend the same arc, they are equal.
Proof:
- Let chord \( AB \) subtend angles \( \angle APB \) and \( \angle AQB \) at points P and Q on the same segment of the circle.
- Join the points to form triangles \( APB \) and \( AQB \).
- Both angles subtend the same arc \( AB \), and lie on the same side of chord AB.
- So, from the geometry of the circle, \( \angle APB = \angle AQB \).
Conclusion: \( \angle \text{in same segment} = \text{equal} \)
Example:
In the circle shown below, find the value of angle \( y \).
▶️ Answer/Explanation
Since both angles subtend the same arc \( AB \), and lie in the same segment, they are equal.
\( y = 47^\circ \)
Reason: Angles in the same segment are equal.
Example:
In the figure below, angle \( \angle APB = 68^\circ \). Find the value of angle \( \angle AQB \).
▶️Answer/Explanation
Both angles subtend the same chord \( AB \) in the same segment.
So, \( \angle AQB = \angle APB = \boxed{68^\circ} \)
5. Opposite Angles of a Cyclic Quadrilateral
A cyclic quadrilateral is a four-sided figure where all vertices lie on the circumference of a circle. In such a quadrilateral, the sum of the opposite angles is always 180°.
Theorem : \( \angle A + \angle C = 180^\circ \quad \text{and} \quad \angle B + \angle D = 180^\circ \)
In any cyclic quadrilateral \( ABCD \), opposite angles are supplementary (i.e., their sum is \( 180^\circ \)).
Proof:
- Let \( ABCD \) be a cyclic quadrilateral with all vertices lying on a circle.
- Angles \( \angle A \) and \( \angle C \) subtend arcs that together form the full circle (i.e., 360°).
- Therefore, \( \angle A + \angle C = 180^\circ \), and similarly, \( \angle B + \angle D = 180^\circ \).
Conclusion: Opposite angles of a cyclic quadrilateral add up to \( 180^\circ \).
Example:
In the cyclic quadrilateral below, find angle \( x \).
▶️ Answer/Explanation
Since opposite angles in a cyclic quadrilateral add up to \( 180^\circ \):
\( x + 80^\circ = 180^\circ \)
\( x = 180^\circ – 80^\circ = \boxed{100^\circ} \)
Example:
Find the value of x in the given figure.
▶️Answer/Explanation
$\text{In } \triangle ACB, $
$\angle ACB = 90^\circ \quad \text{(Angle in a semicircle)} $
$\text{Sum of opposite angles in a cyclic quadrilateral is } 180^\circ$
$\angle ADC + \angle ABC = 180^\circ $
$120^\circ + \angle ABC = 180^\circ $
$\Rightarrow \angle ABC = 60^\circ $
$\text{Now in } \triangle ACB: $
$\angle ACB + \angle CAB + \angle ABC = 180^\circ $
$x + 90^\circ + 60^\circ = 180^\circ $
$x + 150^\circ = 180^\circ$
$x = 180^\circ – 150^\circ $
$x = 30^\circ$
6. Alternate Segment Theorem
The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment of the circle.
Theorem : \( \angle BAT = \angle TBC \)
If a tangent touches a circle at point \( B \), and a chord \( AB \) is drawn, then the angle between the chord and the tangent (e.g. \( \angle BAT \)) is equal to the angle in the alternate segment (e.g. \( \angle CBT \)).
Proof (Conceptual):
- Let a tangent touch the circle at point \( B \), and chord \( AB \) is drawn from that point.
- Let point \( T \) be on the circumference such that triangle \( ABT \) is formed.
- The angle between the tangent and chord (\( \angle BAT \)) is equal to the angle in the alternate segment (\( \angle CBT \)) by this theorem.
- This can be shown using the equality of arcs and inscribed angles and by congruent triangles, though usually taken as a result.
Conclusion: \( \angle \text{between tangent and chord} = \angle \text{in alternate segment} \)
Example:
In the figure, the tangent touches the circle at point \( B \). The angle between chord \( AB \) and the tangent is \( 45^\circ \). Find angle \( ACB \).
▶️ Answer/Explanation
By the Alternate Segment Theorem:
\( \angle ACB = \angle ABT = \boxed{45^\circ} \)
Reason: Angle between tangent and chord equals the angle in the alternate segment.
Example:
In the above diagram, \( \angle BAD = 40^\circ \) and \( \angle BAC = 65^\circ \). What is \( 2x \)?
▶️ Answer/Explanation
According to the tangent-chord theorem, we have:
\( \angle ACB = \angle BAD = 40^\circ \quad \text{(1)} \)
In triangle \( \triangle ABC \), the sum of interior angles is \( 180^\circ \):
\( \angle ACB + \angle BAC + \angle ABC = 180^\circ \)
\( \angle ABC = 180^\circ – \angle ACB – \angle BAC \)
\( \angle ABC = 180^\circ – 40^\circ – 65^\circ \)
\( \angle ABC = 75^\circ \)
\( \angle 2x = \boxed{150^\circ} \)