Coordinate geometry in the (x, y) plane

Q: What is the formula for the distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\)?

\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Q: How do you calculate the midpoint \(M\) of a line segment between points \((x_1, y_1)\) and \((x_2, y_2)\)?

\(M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\)

Q: What is the formula for the gradient \(m\) of a line passing through (x_1, y_1) and (x_2, y_2)?

\(m = \frac{y_2 - y_1}{x_2 - x_1}\)

Q: What is the relationship between the gradients of two perpendicular lines?

The product of their gradients is \(-1\) (\(m_1 \times m_2 = -1\)), or one is the negative reciprocal of the other (\(m_2 = -\frac{1}{m_1}\)).

Q: How do you identify the centre of a circle from its standard equation using the 'Opposite Sign Rule'?

If the terms are \((x - a)\) and \((y - b)\), the centre coordinates are positive \(a\) and \(b\). If the signs in the brackets are positive, the coordinates are negative.

Q: What is the general form of a circle equation and what requirement must the \(x^2\) and \(y^2\) terms meet?

The general form is \(x^2 + y^2 + 2gx + 2fy + c = 0\); the coefficients of \(x^2\) and \(y^2\) must both be 1.

Q: How do you convert a circle equation from general form to standard form?

Use the process of completing the square for both the \(x\) terms and the \(y\) terms separately.

Q: What is the geometric relationship between a tangent and the radius of a circle at the point of contact?

The tangent is always perpendicular (at 90) to the radius at the point of contact.

Q: What are the steps to find the equation of a tangent to a circle at point \(P\)?

1. Find the centre. 2. Find the radius gradient. 3. Find the tangent gradient (negative reciprocal). 4. Use \(y - y_1 = m(x - x_1)\) with point \(P\).

Practice

Mathematics (YMA01) · Pearson Edexcel International A Level · 8 min read · Updated: 7 Sept 2026

Introduction: Stepping up to the Curve!

Welcome to the thrilling world of Coordinate Geometry in the (x, y) plane for Pure Mathematics 2 (P2)!
In P1, you mastered the straight line: calculating distances, finding gradients, and determining midpoints. Now, we are ready to tackle the most perfect curve—the Circle.

Coordinate geometry is vital because it connects algebra (equations) with geometry (shapes). Mastering this chapter will enable you to describe, locate, and analyse geometric figures precisely, setting a strong foundation for calculus and further pure maths topics.

Don't worry if P1 geometry felt challenging! We will break down the essential ideas into easy, manageable steps.

Quick Review: Essential P1 Tools (The Toolkit)

You will constantly rely on these four rules, so make sure they are cemented in your memory:

Distance Formula: The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
Midpoint Formula: The midpoint \(M\) of the line segment is: \[M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\]
Gradient (Slope) Formula: The gradient \(m\) of the line is: \[m = \frac{y_2 - y_1}{x_2 - x_1}\]
Perpendicular Lines: If two lines are perpendicular (meet at 90°), their gradients \(m_1\) and \(m_2\) satisfy: \[m_1 \times m_2 = -1 \quad \text{or} \quad m_2 = -\frac{1}{m_1}\] (This is the negative reciprocal rule—essential for finding tangents!)

Section 1: The Equation of a Circle

A circle is defined as the set of all points that are a fixed distance (the radius, \(r\)) from a fixed point (the centre, \((a, b)\)).

1.1 The Standard Form (Centre-Radius Form)

This is the most straightforward way to write the equation of a circle. It is derived directly from the Distance Formula.

Key Equation: \[(x - a)^2 + (y - b)^2 = r^2\]

\((a, b)\) represents the coordinates of the centre.
\(r\) represents the radius.
\(r^2\) represents the radius squared.

Memory Aid: Finding the Centre

When reading the centre from the standard equation, remember the Opposite Sign Rule.
If the equation is \((x + 5)^2 + (y - 3)^2 = 16\):

For \(x\): \(x + 5 \implies a = -5\)
For \(y\): \(y - 3 \implies b = 3\)
Centre is \((-5, 3)\). Radius \(r = \sqrt{16} = 4\).

1.2 The General Form

Sometimes, the circle equation is expanded out, hiding the centre and radius. This is the General Form:

\[x^2 + y^2 + 2gx + 2fy + c = 0\]

In this form, the coefficients of \(x^2\) and \(y^2\) must both be 1. If they are not (e.g., \(2x^2 + 2y^2 = ...\)), you must divide the entire equation by that coefficient first.

1.3 Converting General Form to Standard Form (Completing the Square)

If you are given the general form, you must use the technique of completing the square (a P1 skill!) to find the centre and radius. This is a common exam requirement.

Step-by-Step: Completing the Square

Example: Find the centre and radius of the circle \(x^2 + y^2 - 6x + 10y - 2 = 0\).

Group the \(x\) terms and \(y\) terms, and move the constant \(c\) to the right side: \[(x^2 - 6x) + (y^2 + 10y) = 2\]
Complete the square for the \(x\) terms: Take half of the coefficient of \(x\) (which is \(-6\)), square it, and add it to both sides.
Half of \(-6\) is \(-3\). Squared is \(9\). \[(x^2 - 6x + 9) + (y^2 + 10y) = 2 + 9\]
Complete the square for the \(y\) terms: Take half of the coefficient of \(y\) (which is \(10\)), square it, and add it to both sides.
Half of \(10\) is \(5\). Squared is \(25\). \[(x^2 - 6x + 9) + (y^2 + 10y + 25) = 2 + 9 + 25\]
Rewrite the bracketed expressions in squared form: \[(x - 3)^2 + (y + 5)^2 = 36\]
Identify the centre and radius:
Centre \((a, b)\) is \((3, -5)\) (Remember the opposite sign!).
Radius \(r = \sqrt{36} = 6\).

Quick Review Box: The Circle

Standard Form: \((x - a)^2 + (y - b)^2 = r^2\)
If you see the General Form, COMPLETE THE SQUARE immediately.
If \(r^2\) is negative, it is not a real circle.

Section 2: Tangents, Normals, and Intersections

The P2 level focuses heavily on the relationship between straight lines and circles, particularly tangents and normals.

2.1 Understanding Tangents and Normals

Imagine a spoke on a bicycle wheel.

A Radius runs from the centre to the edge.
A Tangent is a line that just skims the edge of the circle at a single point, called the point of contact.
The Normal is the line that passes through the centre of the circle and the point of contact. For a circle, the normal is simply the radius (or the line containing the radius).

Crucial Geometric Fact:
A Tangent is always perpendicular (at 90°) to the Radius at the point of contact.

2.2 Finding the Equation of a Tangent

This process is usually the most involved calculation in the chapter.

Step-by-Step: Finding the Tangent Equation

Goal: Find the equation of the tangent to a circle at a given point \(P(x_1, y_1)\).

Find the Centre \((a, b)\) of the circle. (You may need to complete the square first!)
Calculate the Gradient of the Radius (\(m_{radius}\)). Use the gradient formula between the centre \((a, b)\) and the point of contact \(P(x_1, y_1)\). \[m_{radius} = \frac{y_1 - b}{x_1 - a}\]
Calculate the Gradient of the Tangent (\(m_{tangent}\)). Since the tangent is perpendicular to the radius: \[m_{tangent} = -\frac{1}{m_{radius}}\]
Find the Equation of the Tangent Line. Use the point-slope form \(y - y_1 = m_{tangent}(x - x_1)\), using the gradient found in step 3 and the point of contact \(P(x_1, y_1)\).

Note on the Normal: If you need the equation of the Normal, simply use the gradient \(m_{radius}\) from step 2 and the point \(P(x_1, y_1)\). The normal always passes through the centre.

2.3 Lines Intersecting Circles (Simultaneous Equations)

To find the points where a line \(L\) intersects a circle \(C\), you must solve their equations simultaneously.

The simplest way is substitution:

Rearrange the linear equation (the line) to isolate either \(x\) or \(y\) (e.g., \(y = 3x + 1\)).
Substitute this expression into the circle equation.
Expand and simplify the resulting equation. This will always produce a quadratic equation in one variable (either \(x\) or \(y\)).
Solve the quadratic to find the coordinate(s).
Substitute the coordinate(s) back into the linear equation to find the corresponding other coordinate(s).

Using the Discriminant (\(\Delta\))

When the resulting quadratic is in the form \(Ax^2 + Bx + C = 0\), the discriminant, \(\Delta = B^2 - 4AC\), tells you how many intersection points there are without solving the equation fully.

If \(\Delta > 0\): There are two distinct points of intersection (The line is a secant).
If \(\Delta = 0\): There is one point of intersection (The line is a tangent).
If \(\Delta < 0\): There are no real points of intersection (The line misses the circle).

This is extremely useful when the question asks to "show that a line is tangent" or "find the value of k such that the line does not intersect the circle."

Common Pitfall Alert!

When solving simultaneous equations involving circles, ensure you substitute the linear equation into the standard form of the circle (after completing the square) or carefully into the General Form. Do not mix up the variables during substitution!

Key Takeaway for Section 2

The relationship between the radius and the tangent (they are perpendicular) is the foundation for almost every P2 coordinate geometry question. Use the negative reciprocal rule reliably.

Section 3: Diameter and Chord Properties

P2 requires applying geometric properties of lines (like perpendicular bisectors) to chords and diameters of circles.

3.1 The Angle in a Semicircle

Did you know? One of the most beautiful circle theorems is the 'Angle in a Semicircle'.
If \(AB\) is a diameter of a circle, and \(P\) is any other point on the circumference, then the angle \(\angle APB\) is always \(90^\circ\).

How to use this in P2:
If a question gives you three points A, B, and P, and tells you that the angle at P is \(90^\circ\), then A and B must form a diameter.
This means you can calculate:

The Centre: The midpoint of \(AB\).
The Radius: Half the distance of \(AB\).

3.2 Perpendicular Bisectors of Chords

A chord is any straight line segment whose endpoints lie on the circle.
Crucial Geometric Fact: The perpendicular bisector of any chord always passes through the centre of the circle.

Using Chord Properties to Find the Centre

If you are given two points on the circle, \(A\) and \(B\), which form a chord:

Find the Midpoint \(M\) of the chord \(AB\).
Find the Gradient \(m_{AB}\) of the chord \(AB\).
Find the Gradient of the Perpendicular Bisector \(m_{perp}\) (\(m_{perp} = -1/m_{AB}\)).
Find the Equation of the Perpendicular Bisector using \(m_{perp}\) and the midpoint \(M\).

If you repeat this process for a second chord (\(BC\)), the centre of the circle will be the point where the two perpendicular bisector lines intersect (solve the two linear equations simultaneously).

Key Takeaway for Section 3

If you are asked to find the equation of a circle given three points on its circumference, use the perpendicular bisector method twice to locate the centre, and then use the distance formula to find the radius.

Review and Practice Checklist

Before concluding, make sure you can confidently perform these tasks:

Convert between the General Form and the Standard Form of a circle equation using completing the square.
Identify the centre and radius immediately from the Standard Form.
Calculate the gradient of the radius and use the negative reciprocal to find the gradient of the tangent at any given point.
Use the discriminant (\(B^2 - 4AC\)) to determine the relationship between a line and a circle (tangent, secant, or non-intersecting).
Find the centre of a circle given two points that form a diameter (using midpoint).
Find the equation of a circle given three points on its circumference (using two perpendicular bisectors).

Remember, coordinate geometry is all about consistent application of simple formulas. Practice makes perfect! Good luck!

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the formula for the distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\)?

\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

How do you calculate the midpoint \(M\) of a line segment between points \((x_1, y_1)\) and \((x_2, y_2)\)?

\(M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\)

What is the formula for the gradient \(m\) of a line passing through (x_1, y_1) and (x_2, y_2)?

\(m = \frac{y_2 - y_1}{x_2 - x_1}\)

What is the relationship between the gradients of two perpendicular lines?

The product of their gradients is \(-1\) (\(m_1 \times m_2 = -1\)), or one is the negative reciprocal of the other (\(m_2 = -\frac{1}{m_1}\)).

What is the standard (centre-radius) form of the equation of a circle?

\((x - a)^2 + (y - b)^2 = r^2\)

How do you identify the centre of a circle from its standard equation using the 'Opposite Sign Rule'?

If the terms are \((x - a)\) and \((y - b)\), the centre coordinates are positive \(a\) and \(b\). If the signs in the brackets are positive, the coordinates are negative.

What is the general form of a circle equation and what requirement must the \(x^2\) and \(y^2\) terms meet?

The general form is \(x^2 + y^2 + 2gx + 2fy + c = 0\); the coefficients of \(x^2\) and \(y^2\) must both be 1.

How do you convert a circle equation from general form to standard form?

Use the process of completing the square for both the \(x\) terms and the \(y\) terms separately.

What is the geometric relationship between a tangent and the radius of a circle at the point of contact?

The tangent is always perpendicular (at 90) to the radius at the point of contact.

What are the steps to find the equation of a tangent to a circle at point \(P\)?

1. Find the centre. 2. Find the radius gradient. 3. Find the tangent gradient (negative reciprocal). 4. Use \(y - y_1 = m(x - x_1)\) with point \(P\).

How can the discriminant be used to determine if a line is a tangent to a circle?

Substitute the line equation into the circle equation to get a quadratic; if the discriminant \(B^2 - 4AC = 0\), the line is a tangent.

What is the 'Angle in a Semicircle' property?

The angle subtended by a diameter at the circumference is always \(90^\circ\).

What property does the perpendicular bisector of a chord have regarding the circle's centre?

The perpendicular bisector of any chord always passes through the centre of the circle.

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