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Geometry - Circle theorems

Grade 11A Level

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Angles in the same segment: Angles subtended by the same arc (or chord) at the circumference are equal. For example, ∠ACB=∠ADB\angle ACB = \angle ADB because they both stand on arc AB.

Circle showing two angles C and D standing on the same chord AB
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Alternate Segment Theorem: The angle between a tangent and a chord through the point of contact is equal to the angle subtended by the chord in the alternate segment.

Circle with a tangent and an inscribed triangle showing the alternate segment theorem
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Angle in a semi-circle: The angle subtended by a diameter at the circumference is always 90∘90^\circ. This forms a right-angled triangle where the diameter is the hypotenuse.

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Perpendicular from center: A line drawn from the center of a circle perpendicular to a chord bisects the chord. Conversely, the line joining the center to the midpoint of a chord is perpendicular to the chord.

📐Formulae

∠at center=2×∠at circumference\angle \text{at center} = 2 \times \angle \text{at circumference}

∠in semi-circle=90∘\angle \text{in semi-circle} = 90^\circ

∠A+∠C=180∘ and ∠B+∠D=180∘ (for cyclic quadrilateral ABCD)\angle A + \angle C = 180^\circ \text{ and } \angle B + \angle D = 180^\circ \text{ (for cyclic quadrilateral ABCD)}

Radius⊥Tangent⇒θ=90∘\text{Radius} \perp \text{Tangent} \Rightarrow \theta = 90^\circ

PA=PB (where P is an external point and A, B are points of tangency)PA = PB \text{ (where P is an external point and A, B are points of tangency)}

💡Examples

Problem 1:

In a circle with center O, point A and B lie on the circumference. If the angle ∠AOB=110∘\angle AOB = 110^\circ, find the angle ∠ACB\angle ACB where C is a point on the major arc.

Solution:

55∘55^\circ

Explanation:

According to the circle theorem 'Angle at the center is twice the angle at the circumference', ∠ACB=12∠AOB\angle ACB = \frac{1}{2} \angle AOB. Therefore, ∠ACB=110∘/2=55∘\angle ACB = 110^\circ / 2 = 55^\circ.

Problem 2:

ABCD is a cyclic quadrilateral. If ∠ABC=105∘\angle ABC = 105^\circ and ∠BCD=70∘\angle BCD = 70^\circ, find the value of ∠ADC\angle ADC.

Solution:

75∘75^\circ

Explanation:

In a cyclic quadrilateral, opposite angles sum to 180∘180^\circ. The angle opposite to ∠ABC\angle ABC is ∠ADC\angle ADC. Therefore, ∠ADC=180∘−105∘=75∘\angle ADC = 180^\circ - 105^\circ = 75^\circ.

Problem 3:

A tangent PT is drawn from an external point P to a circle with center O and radius 5cm. If the distance PO is 13cm, find the length of the tangent PT.

Solution:

12 cm12\text{ cm}

Explanation:

The radius OT is perpendicular to the tangent PT at point T, forming a right-angled triangle △OTP\triangle OTP. Using Pythagoras' theorem: OT2+PT2=PO2⇒52+PT2=132⇒25+PT2=169⇒PT2=144⇒PT=12 cmOT^2 + PT^2 = PO^2 \Rightarrow 5^2 + PT^2 = 13^2 \Rightarrow 25 + PT^2 = 169 \Rightarrow PT^2 = 144 \Rightarrow PT = 12\text{ cm}.

Problem 4:

In the diagram, O is the center of the circle. Points A, B, and C lie on the circumference. If ∠OBC=35∘\angle OBC = 35^\circ, calculate the size of ∠BAC\angle BAC.

Circle with triangle ABC and center O, showing angle OBC as 35 degrees

Solution:

1. In △OBC,OB=OC (radii of the circle).1. \text{ In } \triangle OBC, OB = OC \text{ (radii of the circle)}. 2. Therefore, △OBC is isosceles, so ∠OCB=∠OBC=35∘.2. \text{ Therefore, } \triangle OBC \text{ is isosceles, so } \angle OCB = \angle OBC = 35^\circ. 3. In △OBC,∠BOC=180∘−(35∘+35∘)=110∘.3. \text{ In } \triangle OBC, \angle BOC = 180^\circ - (35^\circ + 35^\circ) = 110^\circ. 4. Using the center theorem, ∠BAC=12×∠BOC.4. \text{ Using the center theorem, } \angle BAC = \frac{1}{2} \times \angle BOC. 5.∠BAC=110∘2=55∘.5. \angle BAC = \frac{110^\circ}{2} = 55^\circ.

Explanation:

We first identify an isosceles triangle formed by two radii. By finding the angle at the center, we can apply the theorem that the angle at the center is twice the angle at the circumference.

Problem 5:

A cyclic quadrilateral ABCD is inscribed in a circle. Diagonal AC is a diameter. If ∠CAD=40∘\angle CAD = 40^\circ and ∠ABD=x\angle ABD = x, find the value of xx.

Cyclic quadrilateral ABCD with diameter AC, angle CAD=40, angle ABD=x

Solution:

1.∠ADC=90∘ (angle in a semi-circle since AC is a diameter).1. \angle ADC = 90^\circ \text{ (angle in a semi-circle since AC is a diameter)}. 2. In △ADC,∠ACD=180∘−(90∘+40∘)=50∘.2. \text{ In } \triangle ADC, \angle ACD = 180^\circ - (90^\circ + 40^\circ) = 50^\circ. 3.∠ABD and ∠ACD subtend the same arc AD.3. \angle ABD \text{ and } \angle ACD \text{ subtend the same arc AD}. 4. Angles in the same segment are equal, so ∠ABD=∠ACD.4. \text{ Angles in the same segment are equal, so } \angle ABD = \angle ACD. 5.x=50∘.5. x = 50^\circ.

Explanation:

Identify the right angle created by the diameter. Calculate the third angle in triangle ADC, then use the 'angles in the same segment' theorem to relate it to angle x.