A-Level Mathematics October/November 2024 Q10(b): Find the two possible equations of the circle.

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  Let the centre of the circle be C(h, k). From the context of the question (likely part (a)), the centre lies on the perpendicular bisector of the line segment joining (4, 3) and (8, -5). The equation of this line is $x = 2y + 8$. Therefore, the coordinates of the centre can be written in terms of one variable, e.g., C(2k + 8, k).

  The question also implies that the radius of the circle is $r=10$. The general equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$.

  Substituting the centre's coordinates and the radius gives: $(x - (2k+8))^2 + (y - k)^2 = 10^2$

  The circle passes through the point (4, 3). We can substitute these coordinates for x and y to find the value(s) of k. $(4 - (2k+8))^2 + (3 - k)^2 = 100$ $(-2k - 4)^2 + (3 - k)^2 = 100$

  Now, expand the brackets: $(4k^2 + 16k + 16) + (9 - 6k + k^2) = 100$

  Simplify to form a quadratic equation: $5k^2 + 10k + 25 = 100$ $5k^2 + 10k - 75 = 0$

  Divide by 5: $k^2 + 2k - 15 = 0$

  Factorise the quadratic to solve for k: $(k+5)(k-3) = 0$ This gives two possible values for k: $k = 3$ or $k = -5$.

  We find the centre and equation for each case:

  Case 1: $k = 3$ The x-coordinate of the centre is $h = 2(3) + 8 = 6 + 8 = 14$. The centre is (14, 3) and the radius is 10. The equation is $(x - 14)^2 + (y - 3)^2 = 10^2$. $(x - 14)^2 + (y - 3)^2 = 100$

  Case 2: $k = -5$ The x-coordinate of the centre is $h = 2(-5) + 8 = -10 + 8 = -2$. The centre is (-2, -5) and the radius is 10. The equation is $(x - (-2))^2 + (y - (-5))^2 = 10^2$. $(x + 2)^2 + (y + 5)^2 = 100$

  The two possible equations of the circle are $(x - 14)^2 + (y - 3)^2 = 100$ and $(x + 2)^2 + (y + 5)^2 = 100$.

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  How the marks are awarded

  • M1 — For expressing the centre's coordinates in terms of a single variable, like (2k+8, k), using the given line equation.
  • M1 — For substituting a point on the circle, e.g. (4, 3), and the radius ($r=10$) into the circle equation $(x-h)^2+(y-k)^2=r^2$ to form an equation in k.
  • DM1 — For correctly expanding the brackets and simplifying to obtain a 3-term quadratic equation, e.g. $5k^2 + 10k - 75 = 0$.
  • A1 — For correctly solving the quadratic to find the two possible y-coordinates of the centre, $k=3$ and $k=-5$.
  • A1 — For finding both centres and stating the two correct final equations of the circles, ensuring signs and values are correct.

  Common mistakes

  • A student might correctly identify that the centre lies on the line $x=2y+8$ but then try to equate the distances from the general centre (2y+8, y) to the points (4,3) and (8,-5). This simply re-proves that the centre is on the perpendicular bisector and leads to an identity like 0=0, causing them to get stuck.
  • Making algebraic errors when expanding squared terms, particularly $(-2k-4)^2$, often forgetting the middle term or mishandling the negative signs. This leads to an incorrect quadratic and loses all subsequent accuracy marks.
  • Successfully finding the two values for the y-coordinate of the centre ($k=3$ and $k=-5$) but then only finding one of the two possible circle equations, thus losing the final A1 mark.
  • Incorrectly calculating the x-coordinate of the centre after finding k. For instance, for $k=-5$, calculating $h = 2(-5)+8 = -18$ instead of $h=-2$, which results in an incorrect final equation.

  Examiner tip: When a circle's centre lies on a given line and its radius is known, substitute the line's equation into the general circle formula to create a solvable quadratic for one of the centre's coordinates.

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