Higher June 2019 Paper 1 Q21
21.
Diagram not drawn to scale
The diagram shows an arc of a circle with centre O and equation \(x^2 + y^2 = 50\).
P \((5, p)\) lies on the circle.
The arc meets the \(x\)-axis at Q.
The tangent to the circle at P meets the \(x\)-axis at R.
(a) Find the value of \(p\), the \(y\)-coordinate of P. [1]
(b) Show that the equation of the tangent to the circle at P is \(y = mx + 10\), where \(m\) is a constant. [4]
(c) Find the exact length of QR. [2]
| Scheme | Marks | Comments |
|---|---|---|
| \((p =)\ 5\) oe | B1 |
| Scheme | Marks | Comments |
|---|---|---|
| \(\left(m_{\text{radius}} =\right)\ \dfrac{5}{5}\) or 1 | M1 | FT \(\dfrac{\text{'their }p\text{'}}{5}\) |
| \(\left(m_{\tan} =\right)\ -1\) | m1 | FT ‘their 1’ |
| \(5 = -1\times5 + c\) oe | M1 | FT ‘their gradient of tangent’ and ‘their \(p\)’: \(\text{'their }p\text{'} = \text{'their}(-1)\text{'}\times5 + c\) Allow for e.g. \(y - 5 = -1(x - 5)\) or \(\dfrac{y - 5}{x - 5} = -1\) |
| \(y = 10 - x\) oe | A1 | Allow for \(y = -x + c\) \(5 = -1\times5 + c\) \(c = 10\) |
Candidates starting with the answer and working back score 2, 3 or 4 SC marks for:
| Scheme | Marks | Comments |
|---|---|---|
| \(5 = 5m + 10\) or \(\dfrac{5 - 0}{5 - 10} = -1\) leading to \(m = -1\) and \(\left(m_{\text{radius}} = \dfrac{5}{5} =\right)\ 1\) | SC2 | |
| \(1\times-1 = -1\) (therefore \(OP \perp PR\)) | SC1 | |
| Conclusion including tangent is perpendicular to radius and therefore \(y = -x + 10\) is the equation of the tangent (at P). | SC1 | NB For final SC1, the equation \(y = -x + 10\) must be stated at some point. |
| Scheme | Marks | Comments |
|---|---|---|
| \(10 - \sqrt{50}\) or \(10 - 5\sqrt2\) ISW | B2 | or exact equivalent B1 for R(10, 0) or OR = 10 or Q(\(\sqrt{50}\), 0) or OQ = \(\sqrt{50}\) or exact equivalent |
| (7) |