Higher November 2017 Paper 1 Q28
28 \(A\), \(B\) and \(C\) are points on the circle \(\quad x^2 + y^2 = 36 \quad\) as shown.
\(A\) is on the \(y\)-axis.
\(B\) is on the \(x\)-axis.
\(M\) is the midpoint of \(AB\).
\(COM\) is a straight line.
(a) Show that the coordinates of \(A\) are (0, 6) [1 mark]
(b) Work out the coordinates of \(B\). [1 mark]
(c) Show that the equation of the straight line passing through \(C\), \(O\) and \(M\) is \(\quad y = x\) [2 marks]
(d) Work out the coordinates of \(C\).
Give your answers in surd form. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| (\(0^2 +\)) \(6^2 = 36\) or (\(OA =\)) radius = 6 or \(\sqrt{36} = 6\) | B1 | oe |
Additional guidance
| \(0 + 36 = 36\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| (6, 0) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\dfrac{6 - \text{their } 0}{0 - \text{their } 6}\) or \(\dfrac{\text{their } 0 - 6}{\text{their } 6 - 0}\) or \(\dfrac{6}{-6}\) or \(\dfrac{-6}{6}\) or \(-1\) | M1 | gradient \(AB\) |
| gradient \(OM \times\) gradient \(AB = -1\) and gradient \(OM = 1\) (and \(y = x\)) | A1 | must see correct working for M1 |
| Alternative method 2 | ||
| \(\left(\dfrac{6 + 0}{2}, \dfrac{0 + 6}{2}\right)\) or (3, 3) | M1 | coordinates of \(M\) |
| gradient \(OM = 1\) (and \(y = x\)) or (0, 0) and (3, 3) (and \(y = x\)) | A1 | must see correct working for M1 |
| Answer | Mark | Comments |
|---|---|---|
| \(x^2 + x^2 = 36\) or \(2x^2 = 36\) or \(y^2 + y^2 = 36\) or \(2y^2 = 36\) or (–)\(6 \cos 45°\) or (–)\(6 \sin 45°\) | M1 | oe equation |
| (–)\(\sqrt{\dfrac{36}{2}}\) or (–)\(\sqrt{18}\) or (–)\(3\sqrt{2}\) or (–)\(\dfrac{6\sqrt{2}}{2}\) or (–)\(\dfrac{6}{\sqrt{2}}\) | M1 | |
| \(\left(-\sqrt{18}, -\sqrt{18}\right)\) or \(\left(-3\sqrt{2}, -3\sqrt{2}\right)\) or \(\left(-\dfrac{6\sqrt{2}}{2}, -\dfrac{6\sqrt{2}}{2}\right)\) or \(\left(-\dfrac{6}{\sqrt{2}}, -\dfrac{6}{\sqrt{2}}\right)\) | A1 | oe surd form |