A-Level Mathematics October/November 2024 Q10(a): Points A and B have coordinates (4, 3) and (8, -5) respectively. A circle with radius 1…

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  The centre of the circle lies on the perpendicular bisector of the chord AB.

  First, find the gradient of the chord AB. $A(4, 3)$, $B(8, -5)$ Gradient of AB, $m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 - 3}{8 - 4}$ $m_{AB} = \frac{-8}{4} = -2$

  Next, find the midpoint of the chord AB. Midpoint $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{4 + 8}{2}, \frac{3 + (-5)}{2} \right)$ $M = \left( \frac{12}{2}, \frac{-2}{2} \right) = (6, -1)$

  The perpendicular bisector passes through the midpoint M and has a gradient that is the negative reciprocal of the gradient of AB. Gradient of perpendicular bisector, $m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-2} = \frac{1}{2}$.

  Now, find the equation of the perpendicular bisector using the point-slope form $y - y_1 = m(x - x_1)$ with point M(6, -1) and gradient $m = \frac{1}{2}$. $y - (-1) = \frac{1}{2}(x - 6)$ $y + 1 = \frac{1}{2}x - 3$ $y = \frac{1}{2}x - 4$

  Since the centre of the circle must lie on the perpendicular bisector of the chord AB, it must lie on the line with equation $y = \frac{1}{2}x - 4$.

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  How the marks are awarded

  • M1 — Awarded for a correct method to find the gradient of the line segment AB, resulting in -2.
  • M1 — Awarded for a correct method to find the midpoint of AB, resulting in the coordinates (6, -1).
  • DM1 — Awarded for using the negative reciprocal of their AB gradient to find the perpendicular gradient (1/2), and then using this with their midpoint (6, -1) in an attempt to find the equation of the line. This is a dependent method mark.
  • A1 — Awarded for correctly manipulating the equation to the given form y = 1/2 x - 4. As this is an 'Answer Given' question, all working must be shown clearly to earn this mark.

  Common mistakes

  • Calculating the gradient of the perpendicular incorrectly, for example by not taking the negative (using -1/m) and just using the reciprocal (1/m), or by not taking the reciprocal at all.
  • Using one of the original points A or B, instead of the midpoint, to find the equation of the perpendicular bisector.
  • Making an arithmetic error when calculating the midpoint, for example ( (8-4)/2, (-5+3)/2 ).
  • Attempting to solve by setting up two distance equations from a general centre (h, k) to A and B. This is a much longer method and is prone to algebraic errors, though it can lead to the correct answer if executed perfectly.

  Examiner tip: Recognise that the centre of a circle always lies on the perpendicular bisector of any chord, a key geometric property for solving circle problems involving two points.

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