krit.club logo

Geometry - Circle theorems

Grade 11IGCSE

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

🔑Concepts

•

Angle at the center: The angle subtended by an arc at the center is twice the angle subtended at any point on the remaining part of the circumference.

•

Angle in a semi-circle: The angle subtended by a diameter at the circumference is always 90∘90^\circ.

•

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

•

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

•

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

•

Tangents from an external point: Two tangents drawn to a circle from the same external point are equal in length and form congruent triangles with the center.

•

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

📐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}.