Coordinate geometry

• \(m_1 m_2 = -1\); the right-hand side of a circle is \(r^2\); a tangent to a circle comes from the radius; and where two graphs meet is where their equations are solved together.
• \(m_1 m_2 = -1\); centre \((-g, -f)\), radius \(\sqrt{g^2 + f^2 - c}\), and \(r^2\) on the right; the tangent comes from the radius; substitute, solve, back into the line, and read the discriminant.

• 1.3.1 I can find the equation of a straight line given sufficient information
• 1.3.2 I can interpret and use any of the forms y = mx + c, y − y₁ = m(x − x₁), ax + by + c = 0 in solving problems
• 1.3.3 I understand that the equation (x − a)² + (y − b)² = r² represents the circle with centre (a, b) and radius r
• 1.3.4 I can use algebraic methods to solve problems involving lines and circles
• 1.3.5 I understand the relationship between a graph and its associated algebraic equation, and can use the relationship between points of intersection of graphs and solutions of equations

Accuracy for this chapter. Leave exact answers exact: \(2\sqrt{13}\), \(4\sqrt{5}\), \(-5 \pm 5\sqrt{5}\), \(\tfrac{3}{2}\). Give a decimal only when the question asks, or as a bracketed aside, to 3 significant figures. Give a line in the form the question names; if it names none, \(ax + by + c = 0\) with integer coefficients is the safe default. Show the substitution and the quadratic before its roots: an intersection read off a calculator's equation solver, with no working, earns nothing.

• “\((x - 3)^2 + (y + 2)^2 = 25\) has centre \((3, 2)\) and radius 25.” Correct \((x - 3)^2 + (y + 2)^2 = 25\) has centre \((3, -2)\) and radius 5. The centre's coordinates are the numbers that make each bracket zero, so the signs reverse; the right-hand side is \(r^2\), so the radius is \(\sqrt{25}\).
• “To find the tangent to a circle, differentiate its equation.” Correct A tangent to a circle comes from the radius, not from differentiating. The tangent at \(P\) is perpendicular to the radius \(CP\), so its gradient is \(-1 \div (\text{gradient of } CP)\). Differentiating the equation of a circle is implicit differentiation, which the syllabus note to 1.3.4 excludes.
• “The perpendicular to a line of gradient \(\tfrac{2}{3}\) has gradient \(-\tfrac{2}{3}\).” Correct Flip and change the sign: \(-\tfrac{3}{2}\). Check the product: \(\tfrac{2}{3} \times \left(-\tfrac{3}{2}\right) = -1\).
• “\(2x + 3y - 11 = 0\) has gradient 2.” Correct Make \(y\) the subject: \(y = -\tfrac{2}{3}x + \tfrac{11}{3}\), gradient \(-\tfrac{2}{3}\). In general \(ax + by + c = 0\) has gradient \(-\tfrac{a}{b}\).
• “\(x^2 + y^2 - 6x + 4y - 12 = 0\) has centre \((-3, 2)\) and radius \(\sqrt{12}\).” Correct Complete the square: \((x - 3)^2 + (y + 2)^2 = 9 + 4 + 12 = 25\). Centre \((3, -2)\), radius 5. The expanded form's centre is \((-g, -f)\), and the constant term is not \(r^2\).
• “The discriminant is negative, so the line is a tangent.” Correct Negative means the line does not meet the curve. A tangent is the repeated root: discriminant zero.
• “I found \(x\), so I put it into the circle to get \(y\).” Correct Put each \(x\) back into the line. The circle gives two values of \(y\) for most \(x\), and one of them is not on the line.
• “\(m = \dfrac{x_2 - x_1}{y_2 - y_1}\).” Repair Change in \(y\) over change in \(x\): \(\dfrac{y_2 - y_1}{x_2 - x_1}\), subtracting in the same order on top and bottom.
• “Midpoint \(= \left(\dfrac{x_2 - x_1}{2}, \dfrac{y_2 - y_1}{2}\right)\).” Repair The midpoint is the average: \(\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)\). The differences belong in the gradient and the distance.
• “Perpendicular to gradient \(\tfrac{2}{3}\) is gradient \(-\tfrac{2}{3}\).” Repair Flip and change the sign: \(-\tfrac{3}{2}\). Check that \(m_1 m_2 = -1\).
• “\(2x + 3y - 11 = 0\) has gradient \(\tfrac{2}{3}\).” Repair Rearrange: \(y = -\tfrac{2}{3}x + \tfrac{11}{3}\), gradient \(-\tfrac{2}{3}\), which is \(-\tfrac{a}{b}\).
• “The line through \((4, -1)\) and \((4, 6)\) has gradient \(\tfrac{7}{0}\), so it has no equation.” Repair Equal \(x\)-coordinates mean a vertical line: \(x = 4\).
• “\((x - 3)^2 + (y + 2)^2 = 25\) has centre \((3, 2)\) and radius 25.” Repair Centre \((3, -2)\) — the signs reverse — and radius \(\sqrt{25} = 5\).
• “\(x^2 + y^2 + 2gx + 2fy + c = 0\) has centre \((g, f)\).” Repair Centre \((-g, -f)\), radius \(\sqrt{g^2 + f^2 - c}\); or complete the square and read the standard form.
• “\(x^2 + y^2 - 6x + 4y - 12 = 0\), so \(r = \sqrt{12}\).” Repair \(r^2 = g^2 + f^2 - c = 9 + 4 + 12 = 25\), so \(r = 5\). The constant term on its own is not \(r^2\).
• “The circle with centre \((1, -3)\) and radius \(\sqrt{10}\) is \((x - 1)^2 + (y + 3)^2 = \sqrt{10}\).” Repair The right-hand side is \(r^2 = 10\).
• “\(2x^2 + 2y^2 - 8x + 12y - 6 = 0\), so \(g = -4\) and the centre is \((4, -6)\).” Repair Divide by 2 first, so that the \(x^2\) and \(y^2\) coefficients are 1: \(x^2 + y^2 - 4x + 6y - 3 = 0\), centre \((2, -3)\).
• “Differentiating \(x^2 + y^2 = 25\) gives the tangent gradient at \((3, 4)\).” Repair Not in the syllabus for this outcome, and not needed: the tangent at \((3, 4)\) is perpendicular to the radius of gradient \(\tfrac{4}{3}\), so its gradient is \(-\tfrac{3}{4}\).
• “Substituting \(y = 2x + 5\) into \((x - 1)^2 + (y - 2)^2 = 25\): \((x - 1)^2 + (2x + 3)^2 = 25\), and \((2x + 3)^2 = 4x^2 + 9\).” Repair Square the whole bracket: \((2x + 3)^2 = 4x^2 + 12x + 9\). The missing middle term changes every root that follows.
• “The discriminant is negative, so the line is a tangent.” Repair Negative means no intersection. A tangent is the repeated root: discriminant zero.
• “The discriminant of \(x^2 + y^2 - 6x + 4y - 12 = 0\) tells me whether the line meets it.” Repair The discriminant belongs to the quadratic in one variable obtained after substituting the line. Eliminate first.
• “\(x = 3\) or \(x = -1\); in \(x^2 + y^2 = 10\), \(y = \pm 1\) and \(y = \pm 3\), so the line meets the circle at four points.” Repair A line meets a circle at most twice. Put each \(x\) into the line, which gives one \(y\) each.
• “‘The line meets the curve’, so \(b^2 - 4ac > 0\).” Repair “Meets” includes touching: \(b^2 - 4ac \geqslant 0\). The strict \(> 0\) is for “intersects” set against “touches”, or “two distinct points”.
• “The perpendicular bisector of \(AB\) passes through \(A\).” Repair It passes through the midpoint of \(AB\), at right angles to \(AB\).

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• No differentiation here. The syllabus note to this outcome says implicit differentiation is not included, so the gradient of a tangent to a circle is never found by differentiating \(x^2 + y^2 + 2gx + 2fy + c = 0\); it comes from the radius. Tangents and normals to curves such as \(y = x^3 - 2x\), found from \(\dfrac{dy}{dx}\), are chapter 7.
• Interleave with the chapters that use this one. When you reach chapter 7, redo section A's method card: every tangent and normal there ends with \(y - y_1 = m(x - x_1)\). When you reach chapter 10, find the gradient and intercept of a line through two plotted points again. When you reach chapter 17, redo worked example 1's perpendicular bisector and worked example 2's circle, and compare them with the loci there. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: Paper 1 is assumed knowledge for every other paper, and an individual examination question may involve ideas and methods from more than one section of that paper’s content, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Pure Mathematics 1 content, examined in Paper 1. Paper 1 (Pure Mathematics 1) is compulsory for both the AS Level and the A Level: it is 60% of the AS Level and 30% of the A Level, and its content is assumed knowledge for every other paper. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• Section 1.3 content can be set in structured questions such as these, which are the shapes this chapter's mixed challenge practises: a line through given points, with a perpendicular bisector, an intersection or an area to follow; a circle given in the expanded form, whose centre and radius lead to a tangent at a point and then to where that tangent meets an axis; “show that the line is a tangent”; and “find the set of values of \(k\) for which the line meets the curve”. The reasoning sits in the set-up lines: the perpendicular gradient, the completed square, the substitution of the line, and the condition on the discriminant.
• MF19 prints nothing for section 1.3. The gradient, distance and midpoint formulae, the three forms of a line, \(m_1 m_2 = -1\), the equation of a circle in both forms with the centre \((-g, -f)\) and radius \(\sqrt{g^2 + f^2 - c}\), and the three circle facts must all be known. The MF19 card lists them.
• Lengths and constants that come out as surds stay exact: \(2\sqrt{13}\), \(4\sqrt{5}\), \(-5 \pm 5\sqrt{5}\). Give a decimal only when asked, then to 3 significant figures. A line asked for as \(ax + by + c = 0\) needs integer coefficients. Write the substitution of the line into the curve and the resulting quadratic before its roots: intersection points taken from a calculator's equation solver without that working earn nothing.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.