Tangents and chords

Topic summary

To find the equations of tangents and chords related to a circle, we use the geometric properties of circles and lines. Let’s explore how to calculate these step by step.

Finding the Equation of a Tangent to a Circle

The tangent to a circle is a straight line that touches the circle at exactly one point. The tangent is perpendicular to the radius at the point of contact.

Step 1: Start with the Circle Equation and the Point of Contact

Consider the circle with the equation:

\[ x^2 + y^2 = r^2 \]

and a point of contact \( P(x_1, y_1) \) on the circle.

Step 2: Find the Gradient of the Radius

The gradient of the radius from the circle’s centre (at \( (0, 0) \)) to \( P(x_1, y_1) \) is:

\[ \text{Gradient of radius} = \frac{y_1 – 0}{x_1 – 0} = \frac{y_1}{x_1} \]

Step 3: Find the Gradient of the Tangent

The tangent is perpendicular to the radius, so its gradient is the negative reciprocal of the radius gradient:

\[ \text{Gradient of tangent} = -\frac{x_1}{y_1} \]

Step 4: Use the Point-Gradient Formula

The equation of the tangent can be written using the point-gradient formula:

\[ y – y_1 = m(x – x_1) \]

Substitute \( m = -\frac{x_1}{y_1} \):

\[ y – y_1 = -\frac{x_1}{y_1}(x – x_1) \]

Simplify this to get the tangent equation in a standard form.

Finding the Equation of a Chord Between Two Points

A chord is a straight line segment connecting two points on the circle.

Step 1: Start with the Circle Equation and Two Points

Consider the circle with the equation:

\[ x^2 + y^2 = r^2 \]

and two points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) on the circle.

Step 2: Find the Gradient of the Chord

The gradient of the chord is calculated as:

\[ \text{Gradient of chord} = \frac{y_2 – y_1}{x_2 – x_1} \]

Step 3: Use the Point-Gradient Formula

The equation of the chord can be written using the point-gradient formula. Using \( P(x_1, y_1) \) as a reference point:

\[ y – y_1 = m(x – x_1) \]

Substitute \( m = \frac{y_2 – y_1}{x_2 – x_1} \):

\[ y – y_1 = \frac{y_2 – y_1}{x_2 – x_1}(x – x_1) \]

Simplify to find the equation of the chord.

Examples

Tangent Example

Consider the circle \( x^2 + y^2 = 25 \) and the point \( P(3, 4) \). The gradient of the radius is:

\[ \frac{y_1}{x_1} = \frac{4}{3} \]

The tangent’s gradient is:

\[ -\frac{x_1}{y_1} = -\frac{3}{4} \]

Using the point-gradient formula:

\[ y – 4 = -\frac{3}{4}(x – 3) \]

Simplify to get:

\[ 3x + 4y – 25 = 0 \]

Chord Example

Consider the circle \( x^2 + y^2 = 25 \) and points \( P(3, 4) \) and \( Q(4, 3) \). The gradient of the chord is:

\[ \frac{y_2 – y_1}{x_2 – x_1} = \frac{3 – 4}{4 – 3} = -1 \]

Using the point-gradient formula with \( P(3, 4) \):

\[ y – 4 = -1(x – 3) \]

Simplify to get:

\[ x + y – 7 = 0 \]

Summary

• For tangents, use the negative reciprocal of the radius gradient.
• For chords, calculate the gradient between two points and use the point-gradient formula.
• Simplify your equations to standard forms.