Higher June 2019 Paper 6 Q19
19 The point (–5, 2) lies on the circumference of a circle, centre (0, 0).
(a) Find the equation of the circle. [4]
(b) Work out the gradient of the tangent to the circle at (–5, 2). [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(x^2 + y^2 = 29\) oe | 4 | B2 for 29 or \(\sqrt{29}\) or 5.38(5…) to 5.39 or M1 for \(2^2 + 5^2\) or \(\sqrt{2^2 + 5^2}\) or \(2^2 + (-5)^2\) or \(\sqrt{2^2 + (-5)^2}\) AND B1 for \(x^2 + y^2 = k\) where \(k\) is a number \(\gt 0\) or \(x^2 + y^2 = r^2\) | Condone poor use of or missing brackets for M1 e.g. \(-5^2 + 2^2\) or \(2^2 + -5^2\) earns M1, but \(2^2 - 5^2\) does NOT earn M1 Condone other letters instead of \(r\), except \(x\) and \(y\). |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2.5 or \(\frac{5}{2}\) oe | 2 | M1 for \(-\dfrac{2}{5}\) oe or –0.4 seen or use of \(m_1 m_2 = -1\) with their radius gradient | M1 for [\(y =\)] \(\frac{5}{2}x\) [\(+\ c\)] oe Condone \(-\frac{2}{5}x\) seen for M1 |