| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2018 |
| Session | September |
| Marks | 11 |
| Topic | Circles |
| Type | Distance from centre to line |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard circle techniques: completing the square to find centre/radius, checking intersection via discriminant, finding perpendicular line through centre, and calculating distance. All steps are routine applications of well-practiced methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | \(x^2 + 4x^2 + 2x - 32x + 56 = 0\) | M1 |
| \(5x^2 - 30x + 56 = 0\) | ||
| \(30^2 - 4 \times 5 \times 56 = -220\) | M1 | Consider discriminant |
| \(b^2 - 4ac < 0\) hence no real roots so the circle and line do not intersect | A1 | Conclude with no real roots |
Total: [3]
| Answer | Marks | Guidance |
|---|---|---|
| (ii)(a) | Centre of circle is \((-1, 8)\) | B1 |
| Gradient of perpendicular is -0.5 | B1 | For gradient of perpendicular |
| \(y - 8 = -0.5(x + 1)\) | M1 | Attempt equation of line through their circle centre with gradient of -0.5 |
| \(x + 2y = 15\) | A1 | Obtain correct equation (Allow any 3 term equivalent) |
Total: [4]
| Answer | Marks | Guidance |
|---|---|---|
| (ii)(b) | \(5x = 15\) | M1 |
| \(x = 3\), \(y = 6\) | ||
| distance from centre to line is \(\sqrt{4^2 + 2^2} = \sqrt{20} = 2\sqrt{5}\) | M1 | Use Pythagoras to find distance between centre of circle and point of intersection |
| \((x + 1)^2 + (y - 8)^2 = 3^2\) | M1 | Attempt to find radius of circle (Seen at any point in solution – allow back credit to part (a) if the radius is found at the same time as the centre of circle) |
| hence shortest distance between line and circle is \(2\sqrt{5} - 3\) | A1 | Obtain \(2\sqrt{5} - 3\) (Allow any exact equiv) |
Total: [4]
(i) | $x^2 + 4x^2 + 2x - 32x + 56 = 0$ | M1 | Substitute $y = 2x$ into equation of circle and rearrange to three term quadratic
| $5x^2 - 30x + 56 = 0$ | | |
| $30^2 - 4 \times 5 \times 56 = -220$ | M1 | Consider discriminant
| $b^2 - 4ac < 0$ hence no real roots so the circle and line do not intersect | A1 | Conclude with no real roots
Total: [3]
(ii)(a) | Centre of circle is $(-1, 8)$ | B1 | Seen or used
| Gradient of perpendicular is -0.5 | B1 | For gradient of perpendicular
| $y - 8 = -0.5(x + 1)$ | M1 | Attempt equation of line through their circle centre with gradient of -0.5
| $x + 2y = 15$ | A1 | Obtain correct equation (Allow any 3 term equivalent)
Total: [4]
(ii)(b) | $5x = 15$ | M1 | Attempt to solve simultaneously with $y = 2x$
| $x = 3$, $y = 6$ | | |
| distance from centre to line is $\sqrt{4^2 + 2^2} = \sqrt{20} = 2\sqrt{5}$ | M1 | Use Pythagoras to find distance between centre of circle and point of intersection
| $(x + 1)^2 + (y - 8)^2 = 3^2$ | M1 | Attempt to find radius of circle (Seen at any point in solution – allow back credit to part (a) if the radius is found at the same time as the centre of circle)
| hence shortest distance between line and circle is $2\sqrt{5} - 3$ | A1 | Obtain $2\sqrt{5} - 3$ (Allow any exact equiv)
Total: [4]7 A line has equation $y = 2 x$ and a circle has equation $x ^ { 2 } + y ^ { 2 } + 2 x - 16 y + 56 = 0$.
\begin{enumerate}[label=(\roman*)]
\item Show that the line does not meet the circle.
\item (a) Find the equation of the line through the centre of the circle that is perpendicular to the line $y = 2 x$.\\
(b) Hence find the shortest distance between the line $y = 2 x$ and the circle, giving your answer in an exact form.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2018 Q7 [11]}}