Higher June 2022 Paper 5 Q19
19 The graph below shows a circle with centre (0, 0) and equation \(x^2 + y^2 = 169\).
(a) Show that the point (−12, 5) lies on the circumference of the circle. [2]
(b) Find the equation of the tangent to the circle at the point (−12, 5), giving your answer in the form \(y = mx + c\). [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(([-]12)^2 + 5^2\) | 1 | Accept equivalent reasoning e.g. For first mark \(13^2 - ([-]12)^2\) | If \(-12^2 + 5^2\) do not allow first mark |
| 144 + 25 = 169 or \(\sqrt{144 + 25} = 13\) and \(\sqrt{169} = 13\) | 1 | e.g. For second mark 169 – 144 = 25 \(\sqrt{25} = 5\) | If \(\sqrt{169}\) evaluated then it must be 13 For 2 marks there must be no errors leading to the answer |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(y = \frac{12}{5}x + \frac{169}{5}\) final answer | 5 | B4 for answer \(\frac{12}{5}x + \frac{169}{5}\) oe (no \(y =\) ) or correct 3 – term answer in different form e.g. \(5y - 12x = 169\) OR M1 for \(-\frac{5}{12}\) oe and M1 [tangent gradient = ] \(-1 \div \textit{their} -\frac{5}{12}\) oe | Accept e.g. \(y = 2.4x + 33.8\) \(\frac{12}{5}\) oe implies M1M1 unless contradicted |
| AND M1dep for \(y - 5 = \textit{their } \frac{12}{5}(x - (-12))\) oe | Dep on at least M1 oe e.g. \(5 = \textit{their } \frac{12}{5} \times -12 + c\) Do not allow M1 for e.g. grad \(-\frac{5}{12}\) used if the gradient is then changed subsequently to e.g. \(\frac{12}{5}\) | ||
| or M1dep for \(y = \textit{their } \frac{12}{5}x + c\) | Dep on at least M1 Allow ‘\(c\)’ or any value including 0 Answer \(y = \frac{12}{5}x\ [+ c]\) oe implies M1M1M1 | ||