A-Level Mathematics May/June 2024 Q10: The equation of a circle is (x-3)² +y² = 18. The line with equation y = mx+c passes thr…

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  The equation of the circle is $(x-3)^2 + y^2 = 18$. The line has equation $y = mx+c$ and passes through (0, -9).

  First, find the value of c by substituting the point (0, -9) into the line equation: $-9 = m(0) + c \implies c = -9$ So, the equation of the tangent line is $y = mx - 9$.

  To find the points of intersection, substitute the line equation into the circle equation: $(x-3)^2 + (mx-9)^2 = 18$

  Expand the brackets: $(x^2 - 6x + 9) + (m^2x^2 - 18mx + 81) = 18$

  Collect terms to form a quadratic equation in x: $x^2 + m^2x^2 - 6x - 18mx + 9 + 81 - 18 = 0$ $(1+m^2)x^2 - (6+18m)x + 72 = 0$

  Since the line is a tangent to the circle, there is exactly one point of intersection. This means the quadratic equation has one repeated root. Therefore, the discriminant $b^2 - 4ac$ must be zero. Here, $a = (1+m^2)$, $b = -(6+18m)$, and $c = 72$.

  $(-(6+18m))^2 - 4(1+m^2)(72) = 0$ $(6+18m)^2 - 288(1+m^2) = 0$ $(36 + 216m + 324m^2) - (288 + 288m^2) = 0$ $36m^2 + 216m - 252 = 0$

  Divide the equation by 36 to simplify: $m^2 + 6m - 7 = 0$

  Factorise the quadratic to find the possible values of m: $(m+7)(m-1) = 0$ So, the two possible values are $m = 1$ and $m = -7$.

  Now, for each value of m, find the coordinates of the point of tangency by solving the quadratic in x.

  Case 1: $m=1$ Substitute $m=1$ into $(1+m^2)x^2 - (6+18m)x + 72 = 0$: $(1+1^2)x^2 - (6+18(1))x + 72 = 0$ $2x^2 - 24x + 72 = 0$ $x^2 - 12x + 36 = 0$ $(x-6)^2 = 0 \implies x = 6$ Find the corresponding y-coordinate using $y = mx - 9$: $y = (1)(6) - 9 = -3$ The point of tangency is (6, -3).

  Case 2: $m=-7$ Substitute $m=-7$ into $(1+m^2)x^2 - (6+18m)x + 72 = 0$: $(1+(-7)^2)x^2 - (6+18(-7))x + 72 = 0$ $(1+49)x^2 - (6-126)x + 72 = 0$ $50x^2 + 120x + 72 = 0$ Divide by 2: $25x^2 + 60x + 36 = 0$ $(5x+6)^2 = 0 \implies 5x = -6 \implies x = -\frac{6}{5}$ Find the corresponding y-coordinate using $y = mx - 9$: $y = (-7)(-\frac{6}{5}) - 9 = \frac{42}{5} - \frac{45}{5} = -\frac{3}{5}$ The point of tangency is $(-\frac{6}{5}, -\frac{3}{5})$.

  The two possible values of m are 1 and -7. The corresponding points of tangency are (6, -3) and $(-\frac{6}{5}, -\frac{3}{5})$.

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  How the marks are awarded

  • M1 — Finding the equation of the tangent as y = mx - 9 and substituting it into the circle equation to get (x-3)² + (mx-9)² = 18.
  • M1 — Correctly expanding the brackets and collecting all terms on one side to form the quadratic equation (1+m²)x² - (6+18m)$x + 72 = 0$.
  • *M1 — Using the condition for tangency by setting the discriminant b² - 4ac = 0, using the coefficients from the quadratic in x: (-(6+18m))² - 4(1+m²)(72) = 0.
  • DM1 — Simplifying the discriminant equation to a 3-term quadratic in m, i.e., 36m² + $216m - 252 = 0$ or a simplified version like m² + $6m - 7 = 0$.
  • A1 — Correctly solving the quadratic in m to find the two possible values, $m = 1$ and m = -7.
  • DM1 — Substituting $m = 1$ back into the quadratic in x to find the x-coordinate of the first point of tangency, $x = 6$.
  • DM1 — Substituting m = -7 back into the quadratic in x to find the x-coordinate of the second point of tangency, x = -6/5.
  • A1 — Finding both correct coordinate pairs for the points of tangency: (6, -3) and (-6/5, -3/5).

  Common mistakes

  • Making algebraic errors when expanding (mx-9)² or when collecting terms to form the quadratic in x, leading to an incorrect discriminant equation.
  • Forgetting to use the discriminant condition b² - 4ac = 0. Some students may try to solve for x directly, which is not possible without a value for m.
  • Stopping after finding the values of m. The question explicitly asks for the coordinates of the points of tangency, so this would lose the final three marks.
  • Calculation errors when substituting the values of m back to find the x-coordinates, or when finding the corresponding y-coordinates.

  Examiner tip: Remember that a line is tangent to a curve if their simultaneous equations result in a quadratic with exactly one solution, which means the discriminant b² - 4ac must be equal to zero.

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