Mathematics

1. 1.

   A circle has centre (3, -2) and radius 5. Write down its equation and show that the point (6, 2) lies on the circle.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. A circle with centre (a, b) and radius r has equation (x - a)^2 + (y - b)^2 = r^2.
   2. A point is on the circle if it satisfies the equation exactly.

   Marking points

   • The equation is (x - 3)^2 + (y + 2)^2 = 25.
   • Substitutes (6, 2): (6 - 3)^2 + (2 + 2)^2 = 9 + 16.
   • Since 9 + 16 = 25, the point lies on the circle.

   Examiner tip: Watch the signs: centre (3, -2) gives (y + 2), not (y - 2). The right-hand side is r^2, not r.

2. 2.

   The vectors a = (2, -1, 3) and b = (1, 4, -2). Find the scalar product a.b and the angle between a and b, to 1 decimal place.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Multiply corresponding components and add for the scalar product.
   2. Use cos(theta) = (a.b)/(|a||b|); a negative value means the angle is obtuse.

   Marking points

   • a.b = 2 * 1 + (-1) * 4 + 3 * (-2) = -8.
   • |a| = sqrt(14) and |b| = sqrt(21).
   • cos(theta) = -8/sqrt(294), so theta = 117.8 degrees.

   Examiner tip: A negative scalar product means an obtuse angle. Do not drop the minus sign when taking the inverse cosine.

3. 3.

   Points A and B have position vectors (1, 2, 3) and (4, 0, 5). Find a vector equation of the line AB and show that the point C(7, -2, 7) lies on it.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. A line needs a point and a direction; the direction is the difference of the position vectors.
   2. To test a point, equate each component and check the same parameter value works for all three.

   Marking points

   • Direction vector AB = (3, -2, 2).
   • Equation: r = (1, 2, 3) + t(3, -2, 2).
   • Equates components to C: 1 + 3t = 7, 2 - 2t = -2, 3 + 2t = 7.
   • All three give t = 2, so C lies on the line.

   Examiner tip: Check all three components. If one gives a different t, the point is not on the line.

4. 4.

   The lines l1: r = (1, 0, 2) + lambda(1, 1, 1) and l2: r = (2, 3, 2) + mu(1, -1, 2) intersect. Find the coordinates of the point of intersection.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. In three dimensions two lines can be skew, so solve two component equations and then test the third.
   2. Substitute lambda = 2 into l1 (or mu = 1 into l2) to find the point.

   Marking points

   • Equates components: 1 + lambda = 2 + mu, lambda = 3 - mu, 2 + lambda = 2 + 2 mu.
   • Solves the first two equations: mu = 1 and lambda = 2.
   • Checks the third equation: 2 + 2 = 2 + 2 is satisfied, so the lines intersect.
   • Point of intersection is (3, 2, 4).

   Examiner tip: The third equation is the test for intersection. If it fails, the lines are skew, not intersecting.

5. 5.

   The circle C has equation x^2 + y^2 - 6x + 4y - 12 = 0. The line y = x + k is a tangent to C. Find the two possible values of k, giving exact answers.

   [5 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. A tangent is perpendicular to the radius, so its distance from the centre equals the radius.
   2. Write the line as x - y + k = 0 and use the point-to-line distance formula.

   Marking points

   • Completes the square: (x - 3)^2 + (y + 2)^2 = 25.
   • Centre (3, -2) and radius 5.
   • The tangent is at distance 5 from the centre: |3 - (-2) + k|/sqrt(2) = 5.
   • Obtains |5 + k| = 5 sqrt(2).
   • k = -5 + 5 sqrt(2) or k = -5 - 5 sqrt(2).

   Examiner tip: There are two tangents with gradient 1, one on each side of the circle, so expect two values of k.

6. 6.

   A curve has parametric equations x = t^2, y = 2t. Show that the tangent at P(p^2, 2p) has equation py = x + p^2. Hence find where the tangent at P meets the x-axis, and the equation of the tangent when p = 3.

   [5 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. For parametric curves, the gradient is dy/dx = (dy/dt)/(dx/dt).
   2. Use the point-slope form at the parameter value p, then clear fractions.

   Marking points

   • dy/dx = (dy/dt)/(dx/dt) = 2/(2t) = 1/t.
   • At t = p the tangent is y - 2p = (1/p)(x - p^2).
   • Multiplies by p: py - 2p^2 = x - p^2, so py = x + p^2.
   • Puts y = 0: x = -p^2, so it meets the x-axis at (-p^2, 0).
   • When p = 3 the tangent is 3y = x + 9 (meeting the x-axis at (-9, 0)).

   Examiner tip: Replace t by p after differentiating. Keep the general tangent equation before substituting p = 3.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.