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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