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Geometry - Circle Theorems

Grade 10IGCSE

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

🔑Concepts

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Angle at the center is twice the angle at the circumference standing on the same arc.

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Angle in a semicircle is always a right angle (90°).

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Angles in the same segment are equal (angles subtended by the same arc at the circumference).

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Opposite angles of a cyclic quadrilateral sum to 180°.

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The angle between a tangent and the radius at the point of contact is 90°.

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Tangents to a circle from a common external point are equal in length.

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Alternate Segment Theorem: The angle between a tangent and a chord is equal to the angle in the alternate segment.

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A radius that is perpendicular to a chord bisects the chord.

📐Formulae

∠BOC=2×∠BAC\angle BOC = 2 \times \angle BAC (where O is the center)

∠A+∠C=180∘\angle A + \angle C = 180^\circ (for cyclic quadrilateral ABCD)

Arc Length=θ360×2πr\text{Arc Length} = \frac{\theta}{360} \times 2\pi r

Sector Area=θ360×πr2\text{Sector Area} = \frac{\theta}{360} \times \pi r^2

Angle in semicircle=90∘\text{Angle in semicircle} = 90^\circ

💡Examples

Problem 1:

Points A, B, and C lie on a circle with center O. If reflex angle ∠BOC=260∘\angle BOC = 260^\circ, find the angle ∠BAC\angle BAC.

Solution:

∠BAC=50∘\angle BAC = 50^\circ

Explanation:

First, find the interior angle ∠BOC=360∘−260∘=100∘\angle BOC = 360^\circ - 260^\circ = 100^\circ. Using the theorem 'angle at the center is twice the angle at the circumference', ∠BAC=12∠BOC=100∘2=50∘\angle BAC = \frac{1}{2} \angle BOC = \frac{100^\circ}{2} = 50^\circ.

Problem 2:

ABCD is a cyclic quadrilateral. If ∠ABC=115∘\angle ABC = 115^\circ and ∠CAD=30∘\angle CAD = 30^\circ, find ∠ACD\angle ACD if AD is a diameter.

Solution:

∠ACD=65∘\angle ACD = 65^\circ

Explanation:

  1. Opposite angles in a cyclic quad sum to 180∘180^\circ, so ∠ADC=180∘−115∘=65∘\angle ADC = 180^\circ - 115^\circ = 65^\circ. 2. Since AD is a diameter, ∠ACD=90∘\angle ACD = 90^\circ is incorrect as we need to find the specific angle in triangle ACD. Actually, in △ACD\triangle ACD, the angle in a semicircle is ∠ACD=90∘\angle ACD = 90^\circ. Wait, if AD is diameter, then ∠ACD=90∘\angle ACD = 90^\circ by the 'angle in a semicircle' theorem.

Problem 3:

A tangent is drawn from point T to a circle at point P. If O is the center, OT=13OT = 13 cm, and the radius OP=5OP = 5 cm, find the length of the tangent TP.

Solution:

TP=12TP = 12 cm

Explanation:

The radius is perpendicular to the tangent (OP⊥PTOP \perp PT), forming a right-angled triangle OPT. Using Pythagoras' theorem: TP2=OT2−OP2=132−52=169−25=144TP^2 = OT^2 - OP^2 = 13^2 - 5^2 = 169 - 25 = 144. Therefore, TP=144=12TP = \sqrt{144} = 12 cm.