HookHow three circles tell your phone where it is
At midnight on 2 May 2000, the United States government flipped a switch that made every satnav on Earth ten times more accurate overnight. Until then a deliberate error called Selective Availability had been added to civilian GPS signals, scrambling positions by up to 100 metres; switching it off dropped the typical error to under 10 metres and opened the door to the mapping apps now in every pocket. Underneath that convenience is pure coordinate geometry. Each satellite you can see fixes your distance from a known point, which places you somewhere on a circle (in three dimensions, a sphere) of that radius around it. One satellite gives a whole circle of possibilities; two circles cross at a pair of points; a third pins you to a single spot. Finding where circles and lines intersect — the machinery of this section — is quite literally how your phone knows which street you are standing on.
Coordinate geometry turns shape into algebra: a line, a circle or a curving path becomes an equation you can compute with. This section builds that dictionary. You will find the equation of a straight line and use gradients to test whether lines are parallel or perpendicular; write and read the equation of a circle, completing the square to find its centre and radius, and apply the classic circle properties of tangents and chords; and finally handle parametric equations, where a curve is traced out by a third variable, converting between parametric and Cartesian form and using parametrics to model real motion — including a ride on the London Eye.
ModelStraight lines — gradients, parallel and perpendicular
A straight line is fixed by its gradient \(m\) (steepness, \(\dfrac{\text{change in }y}{\text{change in }x}\)) and any one point on it. The workhorse form is \(y-y_1=m(x-x_1)\), which drops straight in a known point \((x_1,\ y_1)\) and gradient; the general tidy form the spec also wants is \(ax+by+c=0\), with integer coefficients and nothing left as a fraction. From two points you get the gradient first, then use either point in the point-gradient form.
The two conditions that unlock most questions are about pairs of lines. Lines are parallel when their gradients are equal, \(m_1=m_2\). Lines are perpendicular when the product of their gradients is \(-1\), so \(m_1m_2=-1\), which means the perpendicular gradient is the negative reciprocal — flip it over and change the sign. That single relationship drives perpendicular bisectors, shortest distances and normals to curves throughout the whole of A-level. Straight lines also model real linear relationships — cost against quantity, distance against time — where the gradient is a rate and the intercept a fixed starting value.
Find the equation of the perpendicular bisector of \(A(1,\ 2)\) and \(B(5,\ 10)\), giving it as \(ax+by+c=0\).
First the gradient of \(AB\): \(m=\dfrac{10-2}{5-1}=\dfrac{8}{4}=2\). The bisector is perpendicular to \(AB\), so its gradient is the negative reciprocal, \(-\dfrac{1}{2}\). It passes through the midpoint of \(AB\): \(\left(\dfrac{1+5}{2},\ \dfrac{2+10}{2}\right)=(3,\ 6)\). Now use \(y-y_1=m(x-x_1)\): \(y-6=-\dfrac{1}{2}(x-3)\). Multiply through by \(2\): \(2y-12=-(x-3)=-x+3\), which rearranges to \(x+2y-15=0\). Check the midpoint satisfies it: \(3+12-15=0\).
ModelThe circle — equation, centre, radius and its properties
A circle of centre \((a,\ b)\) and radius \(r\) has equation \((x-a)^2+(y-b)^2=r^2\) — Pythagoras applied to the horizontal and vertical distances from the centre. Read the centre with the signs flipped from the brackets: \((x-3)^2+(y+2)^2=25\) has centre \((3,\ -2)\), not \((-3,\ 2)\), and radius \(\sqrt{25}=5\). Exam circles usually arrive in expanded form, \(x^2+y^2+\dots=0\), and you recover the centre and radius by completing the square in \(x\) and in \(y\).
Three circle properties turn geometry into quick answers. First, the angle in a semicircle is a right angle — an angle subtended by a diameter is always \(90^\circ\). Second, a tangent meets the circle at one point and is perpendicular to the radius drawn to that point, so the tangent's gradient is the negative reciprocal of the radius's gradient. Third, the perpendicular from the centre to a chord bisects it. Together these let you find tangents, decide whether a line cuts, touches or misses a circle, and locate centres from chords.
A circle has equation \(x^2+y^2-6x+4y-12=0\). Find its centre and radius, then the tangent at the point \((6,\ 2)\).
Complete the square: \(x^2-6x=(x-3)^2-9\) and \(y^2+4y=(y+2)^2-4\), so \((x-3)^2-9+(y+2)^2-4-12=0\), giving \((x-3)^2+(y+2)^2=25\). The centre is \((3,\ -2)\) and the radius is \(5\). Check \((6,\ 2)\) lies on it: \((6-3)^2+(2+2)^2=9+16=25\). The radius to \((6,\ 2)\) has gradient \(\dfrac{2-(-2)}{6-3}=\dfrac{4}{3}\), so the tangent (perpendicular to it) has gradient \(-\dfrac{3}{4}\). Its equation: \(y-2=-\dfrac{3}{4}(x-6)\), which tidies to \(3x+4y-26=0\).
MechanismParametric equations and converting to Cartesian
Sometimes a curve is best described not by a direct link between \(x\) and \(y\), but by letting both depend on a third variable called a parameter, usually \(t\) or \(\theta\). Parametric equations such as \(x=2t,\ y=t^2-1\) trace the curve out as \(t\) runs through its values — think of \(t\) as time and the point \((x,\ y)\) as a particle moving along the path. This is the natural language for motion and for curves that a single \(y=f(x)\) cannot capture cleanly.
To convert to Cartesian form you eliminate the parameter: make \(t\) the subject of the simpler equation and substitute it into the other, leaving a direct relationship between \(x\) and \(y\). When the parameter is an angle, the elimination usually leans on a trig identity — most often \(\cos^2\theta+\sin^2\theta=1\). Going the other way, Cartesian to parametric, you choose a convenient parameter; a circle \(x^2+y^2=r^2\), for instance, is neatly written as \(x=r\cos\theta,\ y=r\sin\theta\). The one habit that matters: never leave an answer in terms of \(t\) when the question asks for a Cartesian equation.
Convert \(x=2t,\ y=t^2-1\) to Cartesian form, then show \(x=3\cos\theta,\ y=3\sin\theta\) is a circle.
For the first, the simpler equation is \(x=2t\), so make \(t\) the subject: \(t=\dfrac{x}{2}\). Substitute into \(y=t^2-1\): \(y=\left(\dfrac{x}{2}\right)^2-1=\dfrac{x^2}{4}-1\). The parameter is gone and the curve is the parabola \(y=\dfrac{x^2}{4}-1\).
For the second, you cannot make \(\theta\) the subject tidily, so use the identity instead. Square both: \(x^2=9\cos^2\theta\) and \(y^2=9\sin^2\theta\). Add them: \(x^2+y^2=9(\cos^2\theta+\sin^2\theta)=9\). That is a circle, centre the origin, radius \(3\).
CaseParametric modelling — a turn on the London Eye
Parametric equations earn their keep in modelling, because so much of the real world is a point moving with time. Projectiles, robot arms, planetary orbits and fairground rides are all cleaner as \(x(t)\) and \(y(t)\) than as a single \(y=f(x)\). The spec asks you to build such a model, use it to answer questions, and — as always in applied maths — judge its limitations.
Take the London Eye, opened on the South Bank in 2000: a wheel of radius about \(60\) m whose centre sits roughly \(68\) m above the ground, turning once every \(30\) minutes. Measure time \(t\) in minutes from the lowest point. A capsule's position is then modelled parametrically by \(x=60\sin\!\left(\dfrac{2\pi t}{30}\right)\) for the horizontal displacement and \(y=68-60\cos\!\left(\dfrac{2\pi t}{30}\right)\) for the height above the ground. The \(\cos\) starts at \(1\), so at \(t=0\) the height is \(68-60=8\) m; half a turn later, at \(t=15\), it is \(68+60=128\) m at the top. The model assumes a constant speed and a perfectly circular path — real limitations worth naming are that the wheel slows to let passengers on and off, and that the boarding platform is a few metres above the ground, so the true minimum is not exactly \(8\) m.
Using \(y=68-60\cos\!\left(\dfrac{2\pi t}{30}\right)\), find a capsule's height \(5\) minutes after boarding.
Substitute \(t=5\): the angle is \(\dfrac{2\pi\times 5}{30}=\dfrac{\pi}{3}\), which is \(60^\circ\). Then \(\cos\dfrac{\pi}{3}=\dfrac{1}{2}\), so \(y=68-60\times\dfrac{1}{2}=68-30=38\) m. After five minutes the capsule is \(38\) m up — a third of the way round the turn but well under a third of the way up, because height climbs slowly near the bottom of a circle and fastest at the sides. That uneven rate is exactly the behaviour a parametric model captures and a straight-line model would miss, which is why the ride is modelled this way.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Coordinate geometry rewards a clear diagram and a fixed order of operations. On line questions, compute the gradient before anything else, remember that a perpendicular gradient is the negative reciprocal (product \(-1\)), and give final answers as \(ax+by+c=0\) with integer coefficients when that form is asked for — quote a fraction-free line. A one-line check that a known point satisfies your equation catches most slips.
For circles, complete the square carefully and read the centre with flipped signs and the radius as the square root of the right-hand side — losing the square root is the most common error on the whole topic. When a tangent is wanted, find the radius gradient to the point of contact, take its negative reciprocal, and use the point-gradient form; when deciding whether a line cuts, touches or misses a circle, substitute the line into the circle equation and use the discriminant of the resulting quadratic. On parametric questions, state explicitly that you are eliminating the parameter, choose the easier equation to rearrange, and reach for \(\cos^2\theta+\sin^2\theta=1\) whenever the parameter is an angle. On a modelling part, always substitute cleanly, keep the units, and finish with one honest limitation of the model — the mark scheme reserves a mark for the sentence that says where the model breaks down.