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.

Image courtesy of BBC

**Example: Equation of a Tangent**

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

**Solution:**

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

- Centre: $(4, -1)$
- Radius: $4$

**2. Gradient of Radius**:

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

**3. Gradient of Tangent**:

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

**4. Equation of Tangent**:

- Point-slope form at $(6, 2.5)$: $y - 2.5 = -0.57(x - 6)$
- Simplify: $y = -0.57x + 5.92$

**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.

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.

**2. Triangle Formation**: Connect $A$ to $C$ and $B$ to $C$, forming triangle $ABC$ with $AB$ 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 $AB$ is the diameter and $C$ is on the circumference, angle (ACB) must be 90 degrees, fulfilling the theorem's criteria.

**Result**: Regardless of where $C$ is on the semicircle, angle $ACB$is 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 $y = x - 2$ intersects the circle $x^2 + y^2 = 25$.

**Solution**:

**1. Substitute **$y$:

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

**2. Solve Quadratic**:

Simplify to $2x^2 - 4x - 21 = 0$ and solve for $x$.

**3. Find **$y$:

Use the found $x$ values in $y = x - 2$ to get $y$ values.

**4. Intersection Points**:

Combine $x$ and $y$ values for intersection points.

**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 $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the centre and $r$ is the radius.

**Example: Finding Centre and Radius**

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

**Solution**:

**1. Complete the Square:**

**2. Centre and Radius**:

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

**Conclusion**: The circle's centre is $(3, -4)$ and its radius is $6$.

**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 $(1, 7)$ to the circle $x^2 + y^2 = 20$.

#### Solution:

**1. Calculate Secant Length**:

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

**2. Apply Tangent-Secant Theorem**:

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

**3. Tangent Length Calculation**:

- Using the theorem, $\text{Tangent Length} = \sqrt{\text{Secant Length}^2 - 20}.$
- Resulting in a tangent length of approximately $5.48$ 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 $5\sqrt{2}$ cm, given the radius is 5 cm.

**Result**: The length of the chord is approximately $7.07$ cm, demonstrating how geometric properties facilitate the calculation of distances in a circle.

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

Image courtesy of Mathmonks

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.