krit.club logo

Geometry - Circles (Angle properties, Cyclic properties, Tangent and Secant properties)

Grade 10ICSE

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

🔑Concepts

•

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. This implies that all angles subtended by the same arc in the same segment are equal.

Diagram showing the angle at the center O (2x) is double the angle at the circumference P (x) for arc AB.
•

In a cyclic quadrilateral, the sum of opposite angles is 180∘180^\circ. Furthermore, the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Cyclic quadrilateral ABCD inscribed in a circle.
•

Angles in the same segment of a circle are equal. This is a direct consequence of the central angle theorem.

Angles ACB and ADB in the same segment are equal.
•

The angle between a tangent and a chord through the point of contact is equal to the angle subtended by the chord in the alternate segment.

Alternate segment theorem showing angle between tangent PQ and chord TA equals angle TBA.

📐Formulae

Angle at Center: ∠AOB=2×∠APB\angle AOB = 2 \times \angle APB

Cyclic Quadrilateral: ∠A+∠C=180∘,∠B+∠D=180∘\angle A + \angle C = 180^\circ, \angle B + \angle D = 180^\circ

Intersecting Chords (Internal): PA×PB=PC×PDPA \times PB = PC \times PD

Intersecting Secants (External): PA×PB=PC×PDPA \times PB = PC \times PD (where PP is the external intersection point)

Tangent-Secant Theorem: PT2=PA×PBPT^2 = PA \times PB (where PTPT is a tangent and PABPAB is a secant)

Length of tangent from point PP at distance dd from center OO with radius rr: PT=d2−r2PT = \sqrt{d^2 - r^2}

💡Examples

Problem 1:

In a circle with center OO, chord ABAB is equal to the radius of the circle. Find the angle subtended by this chord at a point on the major arc.

Solution:

  1. Let the radius of the circle be rr. Given chord AB=rAB = r.
  2. In △OAB\triangle OAB, OA=OB=rOA = OB = r (radii) and AB=rAB = r (given).
  3. Therefore, △OAB\triangle OAB is an equilateral triangle.
  4. This implies the angle at the center ∠AOB=60∘\angle AOB = 60^\circ.
  5. By the property that the angle at the center is double the angle at the circumference: ∠APB=12∠AOB\angle APB = \frac{1}{2} \angle AOB.
  6. ∠APB=12×60∘=30∘\angle APB = \frac{1}{2} \times 60^\circ = 30^\circ.

Explanation:

We first identify the triangle formed by the radii and the chord. Since all sides are equal, we find the central angle, then apply the theorem relating central angles to angles at the circumference.

Problem 2:

From an external point PP, a tangent PTPT and a secant PABPAB are drawn to a circle. If PT=6 cmPT = 6 \text{ cm} and PA=4 cmPA = 4 \text{ cm}, find the length of ABAB.

Solution:

  1. Use the Tangent-Secant Theorem: PT2=PA×PBPT^2 = PA \times PB.
  2. Substitute the known values: 62=4×PB6^2 = 4 \times PB.
  3. 36=4×PB36 = 4 \times PB.
  4. PB=364=9 cmPB = \frac{36}{4} = 9 \text{ cm}.
  5. Since BB lies on the secant line such that PB=PA+ABPB = PA + AB, we have:
  6. 9=4+AB9 = 4 + AB.
  7. AB=9−4=5 cmAB = 9 - 4 = 5 \text{ cm}.

Explanation:

The Tangent-Secant theorem relates the length of the tangent segment to the product of the entire secant segment and its external portion. Solving for the full secant length allows us to subtract the external part to find the chord length.

Problem 3:

In the given figure, OO is the center of the circle. If ∠AOC=130∘\angle AOC = 130^\circ, find ∠ABC\angle ABC where BB is a point on the circumference in the major segment.

Circle with central angle AOC = 130 degrees and angle ABC at circumference.

Solution:

∠ABC=12×∠AOC\angle ABC = \frac{1}{2} \times \angle AOC ∠ABC=12×130∘\angle ABC = \frac{1}{2} \times 130^\circ ∠ABC=65∘\angle ABC = 65^\circ

Explanation:

According to the property that the angle subtended by an arc at the center of a circle is double the angle subtended by it at any point on the remaining part of the circle, the angle at the circumference ∠ABC\angle ABC is half of the central angle ∠AOC\angle AOC.

Problem 4:

In the figure, two chords ABAB and CDCD of a circle intersect at an internal point PP. If AP=4 cmAP = 4 \text{ cm}, PB=6 cmPB = 6 \text{ cm} and CP=3 cmCP = 3 \text{ cm}, find the length of PDPD.

Circle with chords AB and CD intersecting at internal point P.

Solution:

By the Intersecting Chords Theorem: AP×PB=CP×PDAP \times PB = CP \times PD 4×6=3×PD4 \times 6 = 3 \times PD 24=3×PD24 = 3 \times PD PD=243PD = \frac{24}{3} PD=8 cmPD = 8 \text{ cm}

Explanation:

When two chords of a circle intersect internally, the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.