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Geometry - Circle Theorems

Grade 9IGCSE

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

🔑Concepts

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Angle at the Center Theorem: The angle subtended by an arc at the center of a circle is twice the angle subtended by it at any point on the remaining part of the circle.

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Angle in a Semicircle: The angle subtended by a diameter at the circumference is always 90 degrees.

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Angles in the Same Segment: Angles subtended by the same arc (or chord) at the circumference in the same segment are equal.

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Cyclic Quadrilateral: Opposite angles of a cyclic quadrilateral (a four-sided figure where all vertices lie on a circle) sum to 180 degrees.

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Tangent-Radius Theorem: A tangent to a circle is perpendicular to the radius at the point of contact (90 degrees).

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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 in the alternate segment.

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Tangents from an External Point: Two tangents drawn to a circle from the same external point are equal in length.

📐Formulae

Angle at Center=2×Angle at Circumference\text{Angle at Center} = 2 \times \text{Angle at Circumference}

∠A+∠C=180∘ (Opposite angles of cyclic quadrilateral)\angle A + \angle C = 180^\circ \text{ (Opposite angles of cyclic quadrilateral)}

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

Area of Segment=θ360πr2−12r2sin⁡(θ)\text{Area of Segment} = \frac{\theta}{360}\pi r^2 - \frac{1}{2}r^2\sin(\theta)

💡Examples

Problem 1:

Points A, B, and C lie on a circle with center O. If the angle AOC at the center is 130°, find the angle ABC at the circumference.

Solution:

65°

Explanation:

According to the Angle at the Center Theorem, the angle subtended at the center is twice the angle at the circumference. Therefore, ∠ABC=12×∠AOC=1302=65∘\angle ABC = \frac{1}{2} \times \angle AOC = \frac{130}{2} = 65^\circ.

Problem 2:

ABCD is a cyclic quadrilateral. If ∠DAB=115°\angle DAB = 115°, calculate the size of ∠BCD\angle BCD.

Solution:

65°

Explanation:

In a cyclic quadrilateral, opposite angles are supplementary (sum to 180°). Thus, ∠BCD=180∘−115∘=65∘\angle BCD = 180^\circ - 115^\circ = 65^\circ.

Problem 3:

A tangent PT touches a circle at point T. O is the center of the circle. If OT = 5 cm and OP = 13 cm, find the length of the tangent segment PT.

Solution:

12 cm

Explanation:

The radius OT is perpendicular to the tangent PT, forming a right-angled triangle OTP. Using Pythagoras' theorem: PT2+OT2=OP2⇒PT2+52=132⇒PT2=169−25=144PT^2 + OT^2 = OP^2 \Rightarrow PT^2 + 5^2 = 13^2 \Rightarrow PT^2 = 169 - 25 = 144. Therefore, PT=144=12PT = \sqrt{144} = 12 cm.