UK A-Level: Coordinate Geometry

Coordinate geometry is the powerful fusion of algebra and geometry, providing the tools to solve complex spatial problems with precise calculations. For your A-Level studies, mastering this topic unlocks your ability to analyze lines, circles, and curves, forming a foundation for calculus and further pure mathematics. You will learn to describe shapes with equations, find intersections, and even prove classic geometric theorems using algebraic methods.

The Foundation: Lines and Their Equations

The study of coordinate geometry begins with the straight line. The most critical property of a line is its gradient (or slope), defined as the change in $y$ divided by the change in $x$ between any two points $(x_1, y_1)$ and $(x_2, y_2)$: $m = \frac{y_2 - y_1}{x_2 - x_1}$.

From this, you can derive several key forms of the equation of a line:

• Gradient-Intercept Form: $y = mx + c$, where $m$ is the gradient and $c$ is the $y$-intercept.
• Point-Gradient Form: $y - y_1 = m(x - x_1)$, used when you know a point and the gradient.
• General Form: $ax + by + c = 0$, which is often useful for specific calculations, like finding the distance to a point.

A crucial application is finding the tangent and normal to a curve at a given point. The tangent is a straight line that just touches the curve at that point, sharing its instantaneous gradient. The normal is a straight line perpendicular to the tangent at the point of contact. Since perpendicular lines have gradients whose product is $-1$ ($m_1 \times m_2 = -1$), finding one allows you to immediately find the other.

Example: Find the equation of the tangent to the curve $y = x^2$ at the point $(2, 4)$.

1. Differentiate: $\frac{dy}{dx} = 2x$.
2. At $x=2$, the gradient $m = 2 \times 2 = 4$. This is the gradient of the tangent.
3. Use point-gradient form: $y - 4 = 4(x - 2)$.
4. Simplify to: $y = 4x - 4$.

The normal at the same point would have gradient $m_{normal} = -\frac{1}{4}$, giving the equation $y - 4 = -\frac{1}{4}(x - 2)$.

Analysing Circles Algebraically

A circle is defined as the set of all points equidistant from a central point. If a circle has centre $(a, b)$ and radius $r$, its equation in Cartesian form is: $$(x - a)^2 + (y - b)^2 = r^2$$ The general form of a circle's equation is $x^2 + y^2 + 2gx + 2fy + c = 0$. Here, the centre is $(-g, -f)$ and the radius is $\sqrt{g^2 + f^2 - c}$. It is vital to remember that for a real circle, $g^2 + f^2 - c > 0$.

Finding the intersection between a line and a circle involves substituting the linear equation into the circle's equation. This creates a quadratic. The discriminant ($\Delta = b^2 - 4ac$) of this quadratic tells you the nature of the intersection:

• $\Delta > 0$: The line is a secant, cutting the circle at two distinct points.
• $\Delta = 0$: The line is a tangent, touching the circle at exactly one point.
• $\Delta < 0$: The line does not intersect the circle.

This discriminant method is often the most efficient way to prove a line is a tangent to a circle, as it avoids needing to find the actual point of contact first.

Working with Parametric Equations

Sometimes describing a curve with a single $y=f(x)$ equation is difficult or impossible. Parametric equations use a third variable, a parameter (often $t$ or $\theta$), to define both $x$ and $y$ coordinates separately. For example, a curve might be defined as $x = t^2$, $y = 2t$, where $t \in \mathbb{R}$.

The core skill is conversion to Cartesian form, which means eliminating the parameter to find a direct relationship between $x$ and $y$.

1. From $y = 2t$, express $t$ in terms of $y$: $t = \frac{y}{2}$.
2. Substitute this into the equation for $x$: $x = \left(\frac{y}{2}\right)^2 = \frac{y^2}{4}$.
3. Rearrange to the Cartesian form: $y^2 = 4x$.

Another common parametric form describes a circle: $x = a + r\cos\theta$, $y = b + r\sin\theta$. Here, $\theta$ is the parameter. Using the Pythagorean identity $\cos^2\theta + \sin^2\theta \equiv 1$, you can eliminate $\theta$ to retrieve the standard circle equation $(x-a)^2 + (y-b)^2 = r^2$. Parametric forms are especially useful for modelling motion, where the parameter represents time.

Proving Geometric Properties

A powerful application of coordinate geometry is proving classic geometric facts. You assign coordinates to key points (often strategically placing them on the axes to simplify calculations) and use algebraic techniques to demonstrate properties like perpendicularity, midpoint location, or concurrency.

Example Proof: Prove that the angle in a semicircle is a right angle.

1. Place the circle with its centre at the origin $(0,0)$ and radius $r$. Its equation is $x^2 + y^2 = r^2$.
2. Place the endpoints of the diameter on the x-axis at $A(-r, 0)$ and $B(r, 0)$.
3. Let $P(x, y)$ be any general point on the circumference (but not $A$ or $B$), so it satisfies $x^2 + y^2 = r^2$.
4. Find the gradients of $AP$ and $BP$:

• Gradient of $AP$: $m_1 = \frac{y - 0}{x - (-r)} = \frac{y}{x+r}$
• Gradient of $BP$: $m_2 = \frac{y - 0}{x - r} = \frac{y}{x-r}$

1. If $AP$ is perpendicular to $BP$, then $m_1 \times m_2 = -1$. Compute the product:

$$m_1 m_2 = \frac{y}{x+r} \times \frac{y}{x-r} = \frac{y^2}{x^2 - r^2}$$

1. Since $P$ lies on the circle, $y^2 = r^2 - x^2$. Substitute this in:

$$\frac{r^2 - x^2}{x^2 - r^2} = \frac{-(x^2 - r^2)}{x^2 - r^2} = -1$$ This confirms the gradients are negative reciprocals, hence $AP \perp BP$, and $\angle APB = 90^\circ$.

Common Pitfalls

1. Misremembering the Circle Equation: A frequent error is writing $(x-a)^2 + (y-b)^2 = r$ instead of $r^2$. Always remember the radius is squared. Similarly, when converting from general form $x^2 + y^2 + 2gx + 2fy + c = 0$ to centre-radius form, the centre coordinates are the negatives of the $g$ and $f$ coefficients: $(-g, -f)$.

1. Sign Errors with Gradients and Perpendicular Lines: When using $m_1 \times m_2 = -1$, ensure you handle negative signs correctly. If a tangent has a gradient of $-\frac{2}{3}$, the normal's gradient is $\frac{3}{2}$, not $-\frac{3}{2}$. A quick check: the product $(-\frac{2}{3}) \times (\frac{3}{2}) = -1$.

1. Confusing Parametric and Cartesian Forms: Remember that in parametric equations, $x$ and $y$ are both defined in terms of the parameter. You cannot treat $t$ as a spatial coordinate. When eliminating the parameter, your goal is to remove it completely to find a direct $x$-$y$ relationship, often requiring algebraic manipulation or trigonometric identities.

1. Forgetting the Condition for a Real Circle: From the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, the condition $g^2 + f^2 - c > 0$ is essential for a real, non-zero radius. If this expression equals zero, the "circle" is just a single point; if it's negative, there is no real locus.

Summary

• The equation of a straight line can be expressed in multiple forms ($y=mx+c$, point-gradient, general), with the gradient being its defining feature. Tangents and normals are found using derivatives and the perpendicular gradient rule.
• A circle with centre $(a, b)$ and radius $r$ has the equation $(x-a)^2 + (y-b)^2 = r^2$. The intersection of a line and a circle is determined by solving the simultaneous equations and interpreting the discriminant of the resulting quadratic.
• Parametric equations define a curve using a third parameter ($t$, $\theta$). Converting to Cartesian form requires eliminating this parameter through substitution or the use of identities like $\cos^2\theta + \sin^2\theta \equiv 1$.
• Coordinate geometry allows you to prove geometric theorems by strategically assigning coordinates to points and using algebraic calculation to demonstrate properties like perpendicularity.
• Success in this topic hinges on meticulous algebraic manipulation, careful attention to signs (especially with gradients and circle centres), and a clear understanding of the link between geometric conditions and their algebraic equivalents.