Coordinate Geometry of the Circle

• Say it once and it stays: “Brackets flip the sign; the right-hand side is \(r^2\).” Two errors account for most of the wrong answers in this entire topic, and that sentence is both of them.
• The shape all three share. Every one of them opened by converting the given information into a centre and a radius, and every one of them closed with a check that used a different route from the original working. Those two habits — start with the centre and radius, finish with an independent check — are worth more in this topic than any individual formula.
• Mark yourself on the working, not the answer. Five of these six questions can be finished with a correct final line and incomplete working, and the syllabus states plainly that candidates must show all necessary working. In particular: did you verify the point in question 3 before using it, and did you go past the line in question 4?
• The one-minute version. If you have sixty seconds before an examination, recite: brackets flip the sign; the right-hand side is \(r^2\); substitute the line, then take the discriminant; radius is perpendicular to tangent; subtract the circles, then substitute back. That sentence is the chapter.

• Write down the equation of a circle given its centre and radius, and state the centre and radius given the equation in standard form \((x-a)^2+(y-b)^2=r^2\).
• Explain why the standard form is the distance formula squared, and why the number in each bracket appears with the opposite sign in the centre.
• Complete the square in \(x\) and in \(y\) to convert \(x^2+y^2+2gx+2fy+c=0\) into standard form, balancing every constant you introduce.
• Quote and use centre \(=(-g,-f)\) and \(r=\sqrt{g^2+f^2-c}\), and derive both rather than only recalling them.
• Decide whether a given equation represents a real circle, a single point, or no real locus at all, by testing the sign of \(r^2\).
• Test whether a stated point lies inside, on, or outside a given circle.
• Substitute a line into a circle equation to obtain a single quadratic in one variable, correctly and without losing a term.
• Solve that quadratic to find the exact coordinates of every point of intersection.
• Compute \(\Delta=B^2-4AC\) for the resulting quadratic and classify the line as a chord, a tangent, or non-intersecting.
• Use the perpendicular distance \(d\) from the centre to the line as an independent geometric check, matching \(d<r\), \(d=r\) and \(d>r\) to the three discriminant cases.
• Find the value or values of an unknown constant in a line for which that line is a tangent to a given circle.
• Find the length of a chord, and the coordinates of its midpoint, once the intersections are known.
• Verify first that a stated point of contact actually lies on the circle, before using it for anything.
• Find the gradient of the radius to the point of contact, and hence the gradient of the tangent as its negative reciprocal.
• Write the tangent equation in point–gradient form and simplify it to the form \(px+qy=k\).
• Handle the horizontal and vertical boundary cases correctly: at the top and bottom of a circle the tangent is horizontal, at the left and right extremes it is vertical and has no gradient at all.
• Find the equations of both tangents drawn from a point outside the circle by imposing \(\Delta=0\) on a line through that point.
• Find the length of a tangent from an external point using the right angle at the point of contact.
• Solve every one of these without using calculus, which this outcome does not expect.
• Subtract two circle equations written in expanded form to eliminate \(x^2\) and \(y^2\), and obtain the straight line that results.
• Substitute that line back into either circle to obtain the actual points of intersection, rather than stopping at the line.
• State the equation of the common chord, and say clearly when the line you found is not a common chord because the circles never meet.
• Compute the distance \(d\) between the two centres and compare it with \(r_1+r_2\) and \(|r_1-r_2|\).
• Classify any pair of circles as intersecting at two points, externally tangent, internally tangent, separate, one inside the other, concentric, or coincident.
• Find the point of contact when two circles touch, using the fact that it lies on the line joining the centres.

Build your own questions — it is unusually easy here. Pick a centre and a radius, say \((-2,3)\) and \(5\), and write \((x+2)^2+(y-3)^2=25\). Expand it and you have a general-form question whose answer you already know. Pick any point on it, such as \((2,6)\) — check: \(16+9=25\) — and you have a tangent question. Choose a second centre and radius and you have a two-circle question whose configuration you decided before writing it down. Every question you build this way comes with its own answer key, and building them drills exactly the reading that the topic is testing.

• The prerequisite that catches most people is not the hardest one. It is check 1. Completing the square is easy in isolation and easy to rush, and in this chapter it has to be done twice in the same line of working while three constants are moved across an equals sign. If a centre comes out wrong, this is the first place to look for the reason.
• The general form is not in the List of formulas. \(x^2+y^2+2gx+2fy+c=0\) appears in the syllabus only as an example of a form you must be able to work with, and neither centre \(=(-g,-f)\) nor \(r=\sqrt{g^2+f^2-c}\) is printed anywhere in the paper. Both have to come from you, or be re-derived on the spot by completing the square — which is the more reliable route in any case, and the one section B teaches.
• A single habit removes four of the five. Finish every circle question by substituting your answer back into the original equation. A point of intersection must satisfy both equations. A point of contact must satisfy the circle and the tangent. A centre and radius must reproduce the equation you started from when expanded. None of these checks takes long, and each one catches a different member of the list above.
• Know the shortcut, but do the working. Reading \(g\) and \(f\) straight off is fast and is a legitimate method. It is also the single easiest place in this chapter to drop a factor of two: in \(x^2+y^2+8x-6y-11=0\) the value of \(g\) is \(4\), not \(8\), because the coefficient of \(x\) is \(2g\). On a “show that” question, complete the square in full — the working is what is being asked for, and it is also what protects you from that factor of two.
• The discriminant belongs to the quadratic, not to the circle. \(\Delta=B^2-4AC\) can only be computed once you have a single quadratic in a single variable. Reaching for it before substituting — for instance, trying to apply it to \(x^2+y^2-25=0\), which has two variables — produces a number that means nothing. Substitute first, collect into \(Ax^2+Bx+C=0\) second, then and only then compute \(\Delta\).
• How to spot a boundary case before it bites. Compute the radius gradient first and look at it. If it comes out as \(0\), the tangent is vertical, so write \(x=\) the \(x\)-coordinate of the point. If the gradient is undefined because the two \(x\)-coordinates are equal, the tangent is horizontal, so write \(y=\) the \(y\)-coordinate. In both cases the answer is one line and no reciprocal is taken. It is only the middle ground — a genuine non-zero finite gradient — where \(-\dfrac{1}{m}\) applies.
• Two different situations both give “no common points”, and they are not interchangeable. \(d>r_1+r_2\) means the circles are apart from each other; \(d<|r_1-r_2|\) means one is swallowed by the other. A question asking you to describe the configuration wants the distinction, not just the word “none”. Likewise, both tangency cases give one common point, and again the word “externally” or “internally” is part of the answer.
• Expand first, then subtract. The cancellation only happens when both equations are in the same form with coefficient \(1\) on \(x^2\) and \(y^2\). Subtracting \((x-4)^2+y^2=9\) from \(x^2+y^2=25\) while the first is still bracketed is where sign errors breed. Multiply out, move everything to the left so each equation ends \(=0\), then subtract. If either equation has a coefficient other than \(1\) on the squared terms, divide it through first.
• Three checks that catch most of these twenty-two. Substitute every point you produce back into every equation it should satisfy. Expand every completed square back to confirm you recover the original equation. And for any tangent, confirm that the distance from the centre to the line equals the radius. Each is a few seconds of arithmetic, and between them they detect almost every error on this page.

• Both components are compulsory, both are externally assessed, and both can draw on any part of the content, so circle work can appear in either.
• Every candidate takes both components; 0606 is a single, untiered qualification with no Core or Extended route, so there is no easier paper and no harder one. Candidates answer all questions on both papers and must show all necessary working. Grades A* to E are available, and the examination is offered in the June and November series, and in the March series in India, for this cycle. The two assessment objectives, knowledge and understanding of mathematical techniques, and analysing, interpreting and communicating mathematically, each carry 45–55% of every component — which is why a correct answer with no working is worth so much less than it looks.
• More than you would expect. Circle questions are built out of squares, square roots and fractions, and on Paper 1 you cannot evaluate any of them numerically. Three consequences follow:
• Exact surds stay as surds. A radius of \(\sqrt{20}\) is a finished answer, or \(2\sqrt5\) if simplification is asked for. Never convert it to \(4.47\).
• An awkward number is a warning sign. Without a calculator there is no way to rescue a messy result, so if a completed square is producing \(r^2=\tfrac{37}{4}\), go back and check your balancing before going on. An unbalanced constant produces exactly this symptom, and it is far quicker to find the slip than to carry it through four more lines.
• Verification is cheap and worth doing. Substituting a point back into an equation is a handful of small squares. It costs thirty seconds and catches the sign errors that this topic specialises in.