| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Circles |
| Type | Distance from centre to line |
| Difficulty | Moderate -0.3 This is a straightforward application of the perpendicular distance formula from a point to a line to find the radius, then writing the circle equation. It requires knowing that tangent distance equals radius and using the standard formula, but involves no problem-solving insight or multiple complex steps—slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03f Circle properties: angles, chords, tangents |
| Answer | Marks | Guidance |
|---|---|---|
| \(3y + x = 7 \Rightarrow m = -\frac{1}{3}\) | B1 | 2.1 |
| Gradient of line \(l\) through centre perpendicular to given tangent is 3 | B1 FT | 1.2 |
| Equation of \(l\) is \(y + 2 = 3(x - 3)\) | M1\* | 3.1a |
| \(\left.\begin{array}{l}3y+x=7\\y=3x-11\end{array}\right\} x=4,\ y=1\) | M1dep\* | 1.1 |
| \(r^2 = (4-3)^2 + (1+2)^2\) | M1 | 1.1 |
| \((x-3)^2 + (y+2)^2 = 10\) | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-3)^2 + (y+2)^2 = r^2\) | B1 | |
| \(r^2 = (7-3y-3)^2 + (y+2)^2\) | M1\* | |
| \(10y^2 - 20y + (20-r^2) = 0\) | A1 | |
| \((-20)^2 - 4(10)(20-r^2)\) | M1dep\* | |
| \((-20)^2 - 4(10)(20-r^2) = 0 \Rightarrow r^2 = \ldots\) | M1 | |
| \((x-3)^2 + (y+2)^2 = 10\) | A1 |
## Question 6:
$3y + x = 7 \Rightarrow m = -\frac{1}{3}$ | **B1** | 2.1 |
Gradient of line $l$ through centre perpendicular to given tangent is 3 | **B1 FT** | 1.2 | Correctly uses $m_1m_2 = -1$ for their $m$
Equation of $l$ is $y + 2 = 3(x - 3)$ | **M1\*** | 3.1a | Correct equation of form $y + 2 = M(x-3)$ with any non-zero $M$
$\left.\begin{array}{l}3y+x=7\\y=3x-11\end{array}\right\} x=4,\ y=1$ | **M1dep\*** | 1.1 | Solves simultaneous equations to find point of intersection. **M0** if no working shown but allow following **M1** and **A1** if earned
$r^2 = (4-3)^2 + (1+2)^2$ | **M1** | 1.1 | Correct method to find distance (or distance squared) between centre and points of intersection
$(x-3)^2 + (y+2)^2 = 10$ | **A1** | 2.2a |
**Alternative solution:**
$(x-3)^2 + (y+2)^2 = r^2$ | **B1** | | Correct lhs of equation of circle
$r^2 = (7-3y-3)^2 + (y+2)^2$ | **M1\*** | | Substitutes given line into equation of circle. Any equivalent form
$10y^2 - 20y + (20-r^2) = 0$ | **A1** | | Correct equation in $r$ and either $x$ or $y$. Tidied form needed: $\left(10x^2 - 80x + (250-9r^2)=0\right)$
$(-20)^2 - 4(10)(20-r^2)$ | **M1dep\*** | | Correct use of discriminant on their three-term quadratic in either $x$ or $y$
$(-20)^2 - 4(10)(20-r^2) = 0 \Rightarrow r^2 = \ldots$ | **M1** | | Set discriminant equal to zero and solve for $r$ or $r^2$
$(x-3)^2 + (y+2)^2 = 10$ | **A1** | | Correct rhs of equation of circle
**[6 marks]**
---6 In this question you must show detailed reasoning.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{7fc02f90-8f8b-4153-bba1-dc0807124e96-4_650_661_1765_242}
\end{center}
The diagram shows the line $3 y + x = 7$ which is a tangent to a circle with centre $( 3 , - 2 )$.\\
Find an equation for the circle.
\hfill \mbox{\textit{OCR PURE Q6 [6]}}