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Geometry - Circle Theorems

Grade 12IGCSE

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

🔑Concepts

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Angle at the Center: The angle subtended by an arc at the center is twice the angle subtended at the circumference.

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

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Angles in the Same Segment: Angles subtended by the same arc at the circumference are equal.

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Cyclic Quadrilaterals: Opposite angles in a cyclic quadrilateral (a four-sided shape where all vertices touch the circle) sum to 180°.

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

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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

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

(x−h)2+(y−k)2=r2 (Equation of a circle with center (h,k) and radius r)(x - h)^2 + (y - k)^2 = r^2 \text{ (Equation of a circle with center } (h, k) \text{ and radius } r \text{)}

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

∠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)}

💡Examples

Problem 1:

Points A, B, and C lie on the circumference of a circle with center O. If angle BOC = 130°, find the size of angle BAC.

Solution:

65°

Explanation:

By the Angle at the Center Theorem, the angle subtended by an arc at the center (BOC) is twice the angle subtended at the circumference (BAC). Therefore, ∠BAC=1302=65∘\angle BAC = \frac{130}{2} = 65^\circ.

Problem 2:

In a cyclic quadrilateral PQRS, the angle PQR = 115°. Calculate the size of angle PSR.

Solution:

65°

Explanation:

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

Problem 3:

A tangent is drawn from a point T to a circle at point A. If O is the center of the circle, angle OAT is 90°, and angle OTA is 35°, find angle AOT.

Solution:

55°

Explanation:

The radius OA and tangent TA meet at 90°. In the triangle OAT, the sum of angles is 180°. Therefore, ∠AOT=180−90−35=55∘\angle AOT = 180 - 90 - 35 = 55^\circ.

Problem 4:

A chord AB is drawn in a circle. A tangent is drawn at point A. If the angle between the tangent and chord AB is 42°, what is the angle subtended by chord AB in the alternate segment?

Solution:

42°

Explanation:

According to the Alternate Segment Theorem, the angle between a tangent and a chord is equal to the angle subtended by the chord in the alternate segment.