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CIE A-Level Maths Study Notes

1.3.5 Geometrical Properties of Circles

Exploring the geometrical properties of circles offers a fascinating insight into the world of mathematics. These properties are not only pivotal in theoretical mathematics but also have practical applications in various fields. This section is tailored for A-Level students, aiming to provide a comprehensive understanding of the intricate relationship between circles and lines.

Tangent to a Circle

A tangent to a circle is a line that touches the circle at exactly one point, known as the point of tangency. This unique relationship between a tangent and a circle is governed by several important properties:

  • Perpendicularity to Radius: A tangent at any point on a circle is perpendicular to the radius at that point. This is a fundamental property used in many geometrical proofs and problems.
  • Existence of Tangent: For every point on a circle, there exists exactly one tangent.
tangent of a circle

Image courtesy of BBC

Example: Equation of a Tangent

Question: Find the equation of the tangent to the circle with equation (x4)2+(y+1)2=16(x - 4)^2 + (y + 1)^2 = 16 at the point (6,2.5)(6, 2.5).

Solution:

1. Circle Equation: (x4)2+(y+1)2=16(x - 4)^2 + (y + 1)^2 = 16

  • Centre: (4,1)(4, -1)
  • Radius: 44

2. Gradient of Radius:

  • Gradient = 2.5(1)64=1.75\frac{2.5 - (-1)}{6 - 4} = 1.75

3. Gradient of Tangent:

  • Negative reciprocal of radius's gradient: 0.57-0.57

4. Equation of Tangent:

  • Point-slope form at (6,2.5)(6, 2.5): y2.5=0.57(x6)y - 2.5 = -0.57(x - 6)
  • Simplify: y=0.57x+5.92y = -0.57x + 5.92
tangent of a circle

Angle in a Semicircle

The angle in a semicircle is a classic theorem in circle geometry, stating that any angle inscribed in a semicircle is a right angle (90 degrees). This property stems from the fact that the diameter subtends a right angle to any point on the circle's circumference.

Angles in a Semicircle

Image courtesy of BBC

Example Problem: Angle in a Semicircle

Prove that any angle formed at the circumference by a diameter of a semicircle is a right angle.

Solution:

1. Setup: In a semicircle, draw diameter (AB) and select any point (C) on the semicircle's arc.

angle in semicircle

2. Triangle Formation: Connect AA to CC and BB to CC, forming triangle ABCABC with ABAB as the base.

3. Circle Theorem Application: A key theorem states that an angle formed at the circumference by a diameter is always a right angle (90 degrees).

4. Conclusion: Since ABAB is the diameter and CC is on the circumference, angle (ACB) must be 90 degrees, fulfilling the theorem's criteria.

Result: Regardless of where CC is on the semicircle, angle ACBACBis a right angle by the circle theorem. This illustrates a fundamental property of circles, emphasizing the unique relationship between diameters and angles at the circumference.

Algebraic Methods in Circle Geometry

Algebraic methods are crucial in solving problems involving circles and lines. These methods include working with the standard form of a circle's equation and solving equations simultaneously to find intersections.

Example: Intersection of Line and Circle

Find where the line y=x2y = x - 2 intersects the circle x2+y2=25x^2 + y^2 = 25.

Solution:

1. Substitute yy:

Replace yy in the circle's equation with x2x - 2, getting x2+(x2)2=25x^2 + (x - 2)^2 = 25.

2. Solve Quadratic:

Simplify to 2x24x21=02x^2 - 4x - 21 = 0 and solve for xx.

3. Find yy:

Use the found xx values in y=x2y = x - 2 to get yy values.

4. Intersection Points:

Combine xx and yy values for intersection points.

Intersection of a line and circle illustration

Circles and Coordinate Geometry

In coordinate geometry, circles are represented by equations in the Cartesian plane. The standard form of a circle's equation is (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the centre and rr is the radius.

Example: Finding Centre and Radius

Find the centre and radius of the circle with the equation x2+y26x+8y11=0x^2 + y^2 - 6x + 8y - 11 = 0.

Solution:

1. Complete the Square:

x2+y26x+8y11=0x^2 + y^2 - 6x + 8y - 11 = 0x26x+y2+8y=11x^2 - 6x + y^2 + 8y = 11(x26x+9)+(y2+8y+16)=11+9+16(x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16(x3)2+(y+4)2=11+9+16(x - 3)^2 + (y + 4)^2 = 11 + 9 + 16(x3)2+(y+4)2=36(x - 3)^2 + (y + 4)^2 = 36

2. Centre and Radius:

  • Centre: (3,4)(3, -4).
  • Radius: 36=6\sqrt{36} = 6.

Conclusion: The circle's centre is (3,4)(3, -4) and its radius is 66.

Tangents from a Point Outside a Circle

When a tangent is drawn from a point outside a circle, it creates a unique geometrical scenario. The tangents from the external point to the circle are equal in length.

Example: Length of Tangents

Find the length of the tangents drawn from the point (1,7)(1, 7) to the circle x2+y2=20x^2 + y^2 = 20.

Solution:

1. Calculate Secant Length:

  • Distance from (1,7)(1, 7) to the circle's center (0,0)(0, 0) using Euclidean distance: (10)2+(70)2\sqrt{(1-0)^2 + (7-0)^2}.

2. Apply Tangent-Secant Theorem:

  • Theorem states: Tangent Length2=Secant Length2Radius2\text{Tangent Length}^2 = \text{Secant Length}^2 - \text{Radius}^2.
  • Circle's radius from x2+y2=20x^2 + y^2 = 20 is 20\sqrt{20}.

3. Tangent Length Calculation:

  • Using the theorem, Tangent Length=Secant Length220.\text{Tangent Length} = \sqrt{\text{Secant Length}^2 - 20}.
  • Resulting in a tangent length of approximately 5.485.48 units.

Chords and Secants

Chords and secants are line segments associated with circles. A chord is a line segment with both endpoints on the circle, while a secant is a line that intersects the circle at two points.

Example: Chord Length

Calculate the chord length subtending a right angle at the center of a circle with radius 5 cm.

Solution:

  • Triangle Properties: The chord and radius form an isosceles right triangle at the circle's center.
  • Chord as Hypotenuse: The chord is the hypotenuse of this triangle.
  • Chord Length Calculation: Using the Pythagorean theorem, the chord length is 525\sqrt{2} cm, given the radius is 5 cm.

Result: The length of the chord is approximately 7.077.07 cm, demonstrating how geometric properties facilitate the calculation of distances in a circle.

chords and secants illustration

The graphical representation provided offers a detailed visualization of the geometrical properties inherent in circles

Geometrical properties

Image courtesy of Mathmonks

Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
LinkedIn
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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