CIE IGCSE Mathematics (0580) Circle theorems II Study Notes - New Syllabus
CIE IGCSE Mathematics (0580) Circle theorems II Study Notes
LEARNING OBJECTIVE
- Circle Theorems II
Key Concepts:
- Circle Theorems II
Circle Theorems II
Circle Theorems II
These theorems describe geometric symmetry properties of circles involving chords, bisectors, and tangents. They are useful for solving problems involving lengths and construction.
Key Theorems:
- 1. Equal chords are equidistant from the centre
- 2. The perpendicular bisector of a chord passes through the centre
- 3. Tangents from an external point are equal in length
1. Equal Chords Are Equidistant from the Centre
If two chords in a circle are the same length, then their perpendicular distances from the centre of the circle are equal.
Theorem:
Equal chords are equidistant from the centre of the circle.
Proof:
- Let chords AB and CD be equal in a circle with centre O.
- Drop perpendiculars from O to AB and CD. Let the feet of the perpendiculars be M and N.
- In triangles OMA and ONC:
- OA = OC (radii)
- AB = CD (given)
- AM = CN (since both are half of equal chords)
- By RHS congruency, triangles OMA and ONC are congruent.
- Therefore, OM = ON.
Conclusion: Equal chords are the same distance from the centre.
Example:
In a circle, AB and CD are chords of equal length. The distance from the centre to AB is 6 cm. What is the distance from the centre to CD?
▶️ Answer/Explanation
By the theorem, equal chords are equidistant from the centre.
So, the distance to CD is also = 6 cm.
Example:
The figure shows a circle with center \( O \). \( AB = 8~\text{cm} \) and \( OM = ON = 3~\text{cm} \).
Answer the following:
- Find the length of \( CN \).
- Find the radius of the circle.
- Calculate \( \angle OCN \).
▶️ Answer/Explanation
(i) Find the length of CN:
- Given \( OM = ON \), and \( AB = CD = 8~\text{cm} \) (equal chords)
- \( \angle CNO = 90^\circ \), since \( ON \perp CD \) (the perpendicular bisector of a chord passes through the center)
- \( CN = \frac{8}{2} = 4~\text{cm} \)
(ii) Find the radius of the circle:
- Use Pythagoras in triangle \( \triangle CNO \):
\( OC^2 = ON^2 + CN^2 = 3^2 + 4^2 = 9 + 16 = 25 \)
\( OC = \sqrt{25} = 5~\text{cm} \)
(iii) Calculate \( \angle OCN \):
- Use tangent ratio in right triangle \( \triangle CNO \):
\( \tan(\angle OCN) = \frac{ON}{CN} = \frac{3}{4} \)
\( \angle OCN = \tan^{-1}\left( \frac{3}{4} \right) \approx \boxed{36.9^\circ} \)
2. Perpendicular Bisector of a Chord Passes Through the Centre
If you draw a perpendicular bisector of any chord of a circle, it will always pass through the centre of the circle.
Theorem:
The perpendicular bisector of a chord always passes through the centre of the circle.
Proof:
- Let AB be a chord in a circle with centre O.
- Let the line PQ be the perpendicular bisector of AB, intersecting it at point M.
- We show that O lies on line PQ.
- In triangles OMA and OMB:
- OA = OB (radii)
- AM = MB (M is midpoint)
- \( \angle OMA = \angle OMB = 90^\circ \) (perpendicular)
- So, triangles OMA and OMB are congruent (RHS).
Conclusion: The line passes through O, the centre.
Example:
A chord AB is 10 cm long. A line bisects AB at right angles. What does this line represent?
▶️ Answer/Explanation
Since the line bisects AB at 90°, it is the perpendicular bisector.
By the theorem, it passes through the centre of the circle.
3. Tangents from an External Point Are Equal
From a point outside the circle, two tangents drawn to the circle are always equal in length.
Theorem:
Tangents drawn from an external point to a circle are equal in length.
Proof:
- Let P be an external point, and PA and PB be tangents to the circle at A and B.
- Join O to A and B (radii).
- OA ⊥ PA and OB ⊥ PB (radius ⊥ tangent).
- Triangles OAP and OBP are right-angled at A and B respectively.
- OA = OB (radii), OP is common, and \( \angle OAP = \angle OBP = 90^\circ \).
- So, triangles OAP and OBP are congruent (RHS).
- Therefore, \( PA = PB \).
Conclusion: Tangents from a point outside the circle are equal in length.
Example:
From a point P, two tangents are drawn to a circle and touch it at A and B. If PA = 7.2 cm, find the length of PB.
▶️ Answer/Explanation
By the theorem, tangents from an external point are equal.
So, PB = 7.2 cm
Example:
Consider a chord AB of length 9 cm in a circle of radius 5 cm. Tangents at A and B intersect at point C, as shown below.
What are the lengths of the tangents CA and CB?
▶️ Answer/Explanation
Let the center of the circle be \( O \), and let line \( OC \) intersect \( AB \) at point \( D \).
Note that:
- \( \angle ADC = 90^\circ \)
- In triangles \( \triangle OAC \) and \( \triangle ODA \):
- \( \angle OAC = \angle ODA = 90^\circ \)
- \( \angle DOA = \angle COA \) (common)
So, by AA similarity, triangles \( \triangle ODA \sim \triangle OAC \)
Now use known values:
- \( OA = 5 \, \text{cm} \)
- \( AB = 9 \, \text{cm} \Rightarrow AD = \frac{9}{2} = 4.5 \, \text{cm} \)
Use the Pythagorean Theorem in triangle \( \triangle OAD \):
\( OD^2 = OA^2 – AD^2 = 5^2 – (9/2)^2 = 25 – 81/4 = 19/4 \)
\( OD = \sqrt{19/4} = \sqrt{4.75} \, \text{cm} \)
Now apply similarity ratio: \( \frac{OD}{OA} = \frac{AD}{AC} \Rightarrow AC = \frac{OA \times AD}{OD} \)
\( AC = \frac{5 \times 4.5}{\sqrt{4.75}} = \frac{22.5}{\sqrt{4.75}} \approx 10.3 \, \text{cm} \)
Final Answer:
\( \boxed{10.3~\text{cm}} \)
This is the (approximate) length of both tangents CA and CB.