TMUA Coordinate Geometry: Lines & Circles

Coordinate geometry is one of the most dependable sources of marks on the TMUA, which is exactly why it is worth getting fluent. It takes a geometric picture, a line through two points, a circle touching an axis, a tangent at a known angle, and turns it into algebra you can grind out by hand. There are only a handful of standard forms to recognise, and once they are automatic the questions become short. The work is rarely conceptually hard; it is testing whether you can set up the right equation quickly and not slip in the arithmetic. This guide covers what the paper actually asks, straight lines with worked examples, distance and midpoint, circles and their tangents, and the traps that quietly cost marks.

What the TMUA actually examines

Coordinate geometry sits in the AS-level pure content that makes up the bulk of Paper 1, and it shows up in two guises. Sometimes it is the whole question: find the equation of a line, the centre of a circle, the point where a tangent meets an axis. More often it is the setting for something else, a quadratic to solve where a line meets a curve, a length that feeds into a later step, or a reasoning task on Paper 2 where you have to justify why two lines must be perpendicular.

The examiners are not testing exotic theory. They are testing whether you hold the standard forms in working memory and can deploy them without hesitation. A question that gives you two points and asks for the perpendicular bisector is really three small moves stacked together: midpoint, gradient, perpendicular gradient, then the line equation. None of those is difficult on its own, but you have to chain them cleanly and at speed, which is the skill the whole paper rewards. It overlaps heavily with graphs and transformations, since a circle or a line is also a graph, and the algebra is the same algebra you meet everywhere else on the test.

So the aim here is not to learn new mathematics. It is to make the setup instant so all your thinking goes into the question, not the formula.

Straight lines

Everything about lines comes from one idea: the gradient, the constant rate at which $y$ changes as $x$ changes. For two points $(x_1, y_1)$ and $(x_2, y_2)$ the gradient is $m = \frac{y_2 - y_1}{x_2 - x_1}$, the change in $y$ over the change in $x$. Get the subtraction in the same order on top and bottom and the sign looks after itself.

Once you have a gradient and any point on the line, the cleanest equation to reach for is the point-gradient form $y - y_1 = m(x - x_1)$. It drops out of the gradient definition rearranged, and it is almost always faster than guessing the intercept first. You can tidy it into $y = mx + c$ afterwards if the answer options are written that way.

The two conditions you must know cold. Parallel lines have equal gradients, so $m_1 = m_2$. Perpendicular lines have gradients whose product is $-1$, so $m_1 m_2 = -1$, which means each gradient is the negative reciprocal of the other. If a line has gradient $2$, any line perpendicular to it has gradient $-\frac{1}{2}$. This negative-reciprocal flip is the single most useful fact in the topic, and the TMUA leans on it constantly.

A worked example. Find the equation of the line through $(1, 4)$ that is perpendicular to the line $y = 3x - 2$. The given line has gradient $3$, so the perpendicular gradient is the negative reciprocal, $-\frac{1}{3}$. Now use point-gradient form with the point $(1, 4)$: $y - 4 = -\frac{1}{3}(x - 1)$. Multiply out and tidy to $y = -\frac{1}{3}x + \frac{13}{3}$, or clear fractions to $3y = -x + 13$. Three moves, no calculator, and the only place to slip is the arithmetic at the end, which is why you keep it exact.

Another common task is the perpendicular bisector, the line that cuts a segment in half at a right angle. To find the perpendicular bisector of the segment joining $(2, 1)$ and $(6, 9)$, first get the midpoint, $(4, 5)$, then the segment's gradient, $\frac{9 - 1}{6 - 2} = 2$, then flip it to $-\frac{1}{2}$ for the perpendicular, and finally write $y - 5 = -\frac{1}{2}(x - 4)$. The whole thing is just the building blocks chained together, which is the pattern the calculator-free techniques guide drills you to execute without pausing.

Distance and midpoint

Two short formulas handle most length and centre work. The distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$, which is just Pythagoras applied to the horizontal and vertical gaps. The midpoint is the average of the coordinates, $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.

The distance result almost always lands as a surd, and the answer options will be in simplest surd form, so the last step is tidying. The distance from $(1, 2)$ to $(4, 6)$ is $\sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$, a tidy whole number here, but more often you get something like $\sqrt{50} = 5\sqrt{2}$ that needs simplifying before it matches an option. Solid surd handling pays off directly, which is one reason it sits underneath so much of the paper.

Midpoint earns its keep in two recurring jobs: it is the centre of a circle when you are given the endpoints of a diameter, and it is the first step of any perpendicular bisector. Spotting that a question is secretly a midpoint question is half the battle.

Circles and tangents

The standard equation of a circle with centre $(a, b)$ and radius $r$ is $(x-a)^2 + (y-b)^2 = r^2$. Read it straight off: the numbers subtracted inside the brackets give the centre, and the right-hand side is the radius squared, so you take a square root to get $r$. A circle centred at $(3, -2)$ with radius $4$ is $(x-3)^2 + (y+2)^2 = 16$. Watch the signs: a $+2$ inside the bracket means the centre coordinate is $-2$.

The TMUA loves to hand you a circle in expanded form and make you recover the centre and radius by completing the square. Given $x^2 + y^2 - 6x + 4y - 3 = 0$, group the $x$ terms and the $y$ terms and complete the square on each: $(x-3)^2 - 9 + (y+2)^2 - 4 - 3 = 0$, which rearranges to $(x-3)^2 + (y+2)^2 = 16$. So the centre is $(3, -2)$ and the radius is $4$. Completing the square is the workhorse move here, and it is worth practising until it is automatic, because almost every harder circle question opens with it.

Tangents are where circles meet straight-line work, and the key fact is geometric: the tangent to a circle is perpendicular to the radius at the point of contact. That single property unlocks most tangent questions. If you know the centre and the point of contact, the radius gradient comes from the two points, the tangent gradient is its negative reciprocal, and then point-gradient form gives the tangent line. For example, the tangent to a circle centred at the origin at the point $(3, 4)$ has a radius of gradient $\frac{4}{3}$, so the tangent gradient is $-\frac{3}{4}$ and the tangent is $y - 4 = -\frac{3}{4}(x - 3)$.

A second tangent idea worth holding is the distance test: a line is a tangent to a circle exactly when the perpendicular distance from the centre to the line equals the radius. If that distance is greater than $r$ the line misses the circle, and if it is less the line cuts it in two points. The same logic lets you check tangency algebraically by substituting the line into the circle equation: a tangent gives a quadratic with a repeated root, so its discriminant is zero. That link between a repeated root and a touching line is a favourite Paper 2 reasoning hook.

Try one

Reading about it only takes you so far. Here is a real past-paper question built on exactly this machinery. Give it a genuine attempt against the clock before you reveal the solution, and watch how the coordinate setup is the means to an end rather than the whole task:

If that felt slow, it is almost always the setup that cost you, not the idea, and the setup is the easiest part to make automatic with a little focused practice.

Common traps to avoid

A handful of errors account for most of the dropped marks here, and the answer options are written to catch each one.

• Reversing the gradient subtraction. If you take the $y$ difference top-to-bottom but the $x$ difference bottom-to-top, the sign flips and your line tilts the wrong way. Keep both subtractions in the same order.
• Forgetting the negative reciprocal. Using the reciprocal without the minus sign, so a perpendicular to gradient $2$ comes out as $\frac{1}{2}$ instead of $-\frac{1}{2}$. The product must be $-1$.
• Sign error reading a circle centre. A $(x+2)$ bracket means the centre coordinate is $-2$, not $+2$. The form subtracts the centre, so a plus inside means a negative coordinate.
• Confusing $r$ and $r^2$. The right-hand side of the circle equation is the radius squared, so a circle with $(x-a)^2 + (y-b)^2 = 25$ has radius $5$, not $25$. Always take the square root.
• Botching the completed square constant. When you complete the square you add a constant inside the bracket that must be subtracted back outside. Dropping that correction shifts the radius and ruins the answer.

Most of these are speed slips, not understanding gaps, which is encouraging because they vanish with mixed timed practice. The TMUA is demanding mainly on pace, and clean coordinate work is one of the cheapest places to bank time.

How to drill it

Make this automatic the same way you make anything automatic: short, frequent, mixed reps rather than one marathon. Start by getting the forms cold, writing the gradient, line, distance, midpoint and circle formulas from memory until there is no hesitation, then run quick drills, find the perpendicular through a point, complete the square on a circle, get a tangent, with the clock running.

Then fold it into everything else. Because lines and circles share their algebra with graphs and transformations and with the quadratics you meet across the paper, every line-meets-curve question is secretly extra coordinate practice. A structured plan that builds this in from the start is laid out in how to prepare for the TMUA. Get the setups instant early and a whole band of the paper turns into easy, reliable marks.

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