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Geometry - Circle Theorems

Grade 12A Level

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

🔑Concepts

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The angle subtended by an arc at the center is twice the angle subtended by the same arc at the circumference. If the angle at the circumference is xx, the angle at the center is 2x2x.

Diagram showing the angle at the center (2x) is twice the angle at the circumference (x).
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Angles in the same segment of a circle are equal. This means angles subtended by the same arc at the circumference are identical.

Diagram showing two equal angles in the same segment subtended by the same chord.
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The angle in a semi-circle is always a right angle (90∘90^\circ). Any triangle formed using the diameter as one side and a third point on the circumference is a right-angled triangle.

Triangle in a semi-circle showing a 90 degree angle at the circumference.
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A tangent to a circle is perpendicular to the radius at the point of contact. The angle between the tangent and radius is 90∘90^\circ.

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Opposite angles in a cyclic quadrilateral sum to 180∘180^\circ. If the quadrilateral's vertices all lie on the circle, then A+C=180∘A + C = 180^\circ and B+D=180∘B + D = 180^\circ.

📐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.

Problem 5:

In the circle with center OO, points AA and BB lie on the circumference. The tangent at point AA meets the line OBOB extended at point TT. If ∠AOT=58∘\angle AOT = 58^\circ, calculate the size of ∠ATO\angle ATO.

Right-angled triangle OAT formed by radius OA and tangent AT.

Solution:

1. Tangent AT⊥OA  ⟹  ∠OAT=90∘1. \text{ Tangent } AT \perp OA \implies \angle OAT = 90^\circ 2. In △OAT, sum of angles =180∘2. \text{ In } \triangle OAT, \text{ sum of angles } = 180^\circ 3.∠ATO=180∘−90∘−58∘3. \angle ATO = 180^\circ - 90^\circ - 58^\circ 4.∠ATO=32∘4. \angle ATO = 32^\circ

Explanation:

Since ATAT is a tangent to the circle at point AA, the angle between the radius OAOA and the tangent is 90∘90^\circ. We then use the sum of angles in a triangle to find the remaining angle.

Problem 6:

A circle has a diameter ADAD. Point BB and CC lie on the circumference such that ABCDABCD is a quadrilateral. If ∠CAD=35∘\angle CAD = 35^\circ, find ∠ACD\angle ACD.

Triangle ACD inside a circle where AD is the diameter.

Solution:

1. Since AD is a diameter, ∠ACD is an angle in a semi-circle.1. \text{ Since } AD \text{ is a diameter, } \angle ACD \text{ is an angle in a semi-circle.} 2.∠ACD=90∘2. \angle ACD = 90^\circ

Explanation:

By the circle theorem 'Angle in a semi-circle is 90 degrees', any angle subtended by the diameter at the circumference is a right angle. Since ADAD is the diameter, ∠ACD\angle ACD must be 90∘90^\circ.