| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Circles |
| Type | Line-circle intersection points |
| Difficulty | Moderate -0.3 This is a straightforward multi-part circle question requiring standard techniques: completing the square to find centre/radius, substituting a linear equation into the circle equation to find intersection points, and interpreting the result. All parts are routine A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
5 The equation of a circle is $x ^ { 2 } + y ^ { 2 } + 6 x - 2 y - 10 = 0$.\\
(i) Find the centre and radius of the circle.\\
(ii) Find the coordinates of any points where the line $y = 2 x - 3$ meets the circle $x ^ { 2 } + y ^ { 2 } + 6 x - 2 y - 10 = 0$.\\
(iii) State what can be deduced from the answer to part (ii) about the line $y = 2 x - 3$ and the circle $x ^ { 2 } + y ^ { 2 } + 6 x - 2 y - 10 = 0$.
\hfill \mbox{\textit{OCR H240/01 2018 Q5 [8]}}