A-Level Mathematics May/June 2025 Q9(b): It is now given that k = 10. A circle passes through the points P, Q and R. Find the eq…

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  The centre of the circle, C, is given or found in part (a) to be $(-4, 3)$.

  It is given that $k=10$, so the point R on the circle has coordinates $(2, 10)$.

  To find the equation of the tangent at R, we first need the gradient of the radius connecting the centre C to the point R.

  Gradient of radius CR, $m_{CR} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{10 - 3}{2 - (-4)} = \frac{7}{6}$.

  The tangent to the circle at R is perpendicular to the radius CR. Therefore, the gradient of the tangent, $m_{tangent}$, is the negative reciprocal of the gradient of the radius.

  $m_{tangent} = -\frac{1}{m_{CR}} = -\frac{1}{7/6} = -\frac{6}{7}$.

  Now, we use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, with the point $R(2, 10)$ and the gradient $m_{tangent} = -\frac{6}{7}$.

  $y - 10 = -\frac{6}{7}(x - 2)$

  To write the equation in the form $ax+by+c=0$ where a, b, and c are integers, we multiply the entire equation by 7 to eliminate the fraction:

  $7(y - 10) = -6(x - 2)$

  $7y - 70 = -6x + 12$

  Finally, rearrange all terms to one side of the equation:

  $6x + 7y - 70 - 12 = 0$

  $6x + 7y - 82 = 0$

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  How the marks are awarded

  • B1 — Awarded for correctly stating or using the centre of the circle as $(-4, 3)$, which is seen or implied in the gradient calculation.
  • M1 — Awarded for a correct method to find the gradient of the radius CR by using the coordinates of the centre and the point R, leading to $\frac{7}{6}$.
  • DM1 — Awarded for a correct method to find the equation of the tangent. This involves finding the negative reciprocal of the radius gradient (to get $-\frac{6}{7}$) and using it with the point R $(2, 10)$.
  • A1 — Awarded for a correct equation of the tangent in any form, for example, $y - 10 = -\frac{6}{7}(x - 2)$.
  • A1 — Awarded for the final correct equation rearranged into the required form $ax+by+c=0$ with integer coefficients, such as $6x + 7y - 82 = 0$.

  Common mistakes

  • Using the gradient of the radius for the tangent, instead of the negative reciprocal. This would lead to an incorrect tangent gradient of $\frac{7}{6}$.
  • Using the coordinates of the centre $(-4, 3)$ instead of the point of tangency R $(2, 10)$ when forming the equation of the tangent line.
  • Making an arithmetic error when calculating the negative reciprocal, for example, by only changing the sign to get $-\frac{7}{6}$ instead of $-\frac{6}{7}$.
  • Failing to rearrange the final answer into the required form $ax+by+c=0$ with integer coefficients, leaving it as $y = -\frac{6}{7}x + \frac{82}{7}$ or with fractional terms.

  Examiner tip: Remember that the tangent to a circle at a point is always perpendicular to the radius at that same point, so their gradients are negative reciprocals of each other.

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