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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 of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle. Visually, if you have an arc ABAB and a center OO, drawing lines OAOA and OBOB creates a central angle AOB\angle AOB. If you pick any point PP on the circle's circumference and connect PAPA and PBPB, the resulting angle APB\angle APB will satisfy the relationship AOB=2APB\angle AOB = 2\angle APB.

Angles in the same segment of a circle are equal. This can be visualized by drawing a chord ABAB and picking two points CC and DD on the same side of the chord on the circumference. The angles ACB\angle ACB and ADB\angle ADB will be identical because they are subtended by the same arc ABAB.

The angle in a semi-circle is a right angle (9090^\circ). If you draw a diameter ABAB passing through the center OO and pick any point PP on the circumference, the triangle formed by A,B,A, B, and PP will always have APB=90\angle APB = 90^\circ.

In a cyclic quadrilateral, the opposite angles are supplementary, meaning they add up to 180180^\circ. Visually, if a quadrilateral ABCDABCD has all four vertices lying on the circumference of a circle, then A+C=180\angle A + \angle C = 180^\circ and B+D=180\angle B + \angle D = 180^\circ. Additionally, any exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

The tangent at any point of a circle is perpendicular to the radius through the point of contact. If a line touches the circle at point PP and OO is the center, the radius OPOP and the tangent line meet to form a 9090^\circ angle (OPOP \perp tangent).

From any external point, exactly two tangents can be drawn to a circle, and these tangents are equal in length. If point PP is outside the circle and PAPA and PBPB are the tangents touching the circle at AA and BB respectively, then PA=PBPA = PB. Also, the line OPOP bisects the angle APB\angle APB and AOB\angle AOB.

The Alternate Segment Theorem states that the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. If a tangent XYXY touches the circle at point AA and a chord ABAB is drawn, the angle BAX\angle BAX is equal to the angle subtended by chord ABAB at any point CC in the major segment (BAX=BCA\angle BAX = \angle BCA).

Intersecting Chords and Secant Properties: When two chords ABAB and CDCD intersect inside a circle at point PP, the product of the lengths of their segments is equal (PA×PB=PC×PDPA \times PB = PC \times PD). If the chords intersect externally (secants), the same product rule applies to the segments measured from the external point PP to the circle's boundaries.

📐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=d2r2PT = \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=12AOB\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=94=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.