A-Level Mathematics May/June 2024 Q8: A circle with equation x² + y²-6x+2y-15 = 0 meets the y-axis at the points A and B. The…

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  Step 1: Find the coordinates of A and B

  The circle meets the y-axis where $x=0$. Substitute $x=0$ into the circle's equation: $x^2 + y^2 - 6x + 2y - 15 = 0$ $0^2 + y^2 - 6(0) + 2y - 15 = 0$ $y^2 + 2y - 15 = 0$ Factorising the quadratic equation: $(y+5)(y-3) = 0$ So, the y-coordinates of the intersection points are $y=-5$ and $y=3$. Let the points be $A(0, 3)$ and $B(0, -5)$.

  Step 2: Find the centre of the circle

  Complete the square for the equation $x^2 + y^2 - 6x + 2y - 15 = 0$. $(x^2 - 6x) + (y^2 + 2y) - 15 = 0$ $(x-3)^2 - 9 + (y+1)^2 - 1 - 15 = 0$ $(x-3)^2 + (y+1)^2 = 25$ The centre of the circle, C, is $(3, -1)$.

  Step 3: Find the gradient of the radius and the tangent at A

  The radius connects the centre $C(3, -1)$ to the point $A(0, 3)$. Gradient of radius $CA = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-1)}{0 - 3} = \frac{4}{-3} = -\frac{4}{3}$. The tangent at A is perpendicular to the radius CA. Gradient of tangent at A = $-\frac{1}{\text{gradient of CA}} = -\frac{1}{-4/3} = \frac{3}{4}$.

  Step 4: Find the equation of the tangent at A

  Using the point-gradient form $y - y_1 = m(x - x_1)$ with point $A(0, 3)$ and gradient $m = \frac{3}{4}$: $y - 3 = \frac{3}{4}(x - 0)$ $y = \frac{3}{4}x + 3$

  Step 5: Find the coordinates of P

  Due to the symmetry of the circle and the points A and B being on the y-axis, the intersection point P of the tangents must lie on the horizontal line passing through the centre of the circle. The y-coordinate of the centre is -1, so the y-coordinate of P is also -1. Substitute $y=-1$ into the equation of the tangent at A to find the x-coordinate of P: $-1 = \frac{3}{4}x + 3$ $-4 = \frac{3}{4}x$ $-16 = 3x$ $x = -\frac{16}{3}$

  Therefore, the coordinates of P are $(-\frac{16}{3}, -1)$.

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  How the marks are awarded

  • M1 — Substituting x=0 into the circle equation to get the three-term quadratic y² + 2y - 15 = 0.
  • A1 — Correctly solving the quadratic to find the y-intercepts y = -5 and y = 3.
  • B1 — Correctly completing the square to state or imply the centre of the circle is (3, -1).
  • M1 — Attempting to find the gradient of a radius by using the centre (3, -1) and one of the y-intercepts, e.g., (0, 3).
  • A1 — Correctly calculating the gradient of one of the radii as -4/3 (or 4/3 for the other radius).
  • A1FT — Correctly finding the negative reciprocal of the radius gradient to get the tangent gradient, 3/4. This is a follow-through mark.
  • DM1 — Using the symmetry of the problem by substituting the y-coordinate of the centre (y=-1) into a tangent equation to find the x-coordinate of P. This is dependent on the previous M mark.
  • A1 — Stating the final, correct coordinates of P as (-16/3, -1).

  Common mistakes

  • Errors in completing the square, such as sign mistakes, leading to an incorrect centre like (-3, 1) or (3, 1).
  • Confusing the gradient of the radius with the gradient of the tangent, and using m = -4/3 instead of m = 3/4 for the tangent equation.
  • Making algebraic errors when solving for the coordinates of P, for example, miscalculating -1 - 3 = -4 when solving for x.
  • Finding the x-intercepts instead of the y-intercepts by incorrectly setting y=0 at the start.

  Examiner tip: Recognize and use symmetry in circle problems, as the point where tangents from symmetric points meet will lie on the axis of symmetry.

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