| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Circles |
| Type | Circle through three points using perpendicular bisectors |
| Difficulty | Standard +0.3 This is a standard circle-through-three-points question requiring perpendicular bisector calculation and solving simultaneous equations to find the centre. While it involves multiple steps (finding midpoint, gradient, perpendicular gradient, then either a second bisector or substitution), these are all routine A-level techniques with no novel insight required. Slightly above average difficulty due to the coordinate arithmetic involved, but remains a textbook exercise. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Midpoint \(AB\) is \((3.5, 5.5)\); Gradient \(AB = -\frac{1}{7}\) | B1 | Both. Allow midpoint \(= \left(\frac{0+7}{2}, \frac{6+5}{2}\right)\) ISW |
| Gradient of perpendicular bisector \(-1/\left(-\frac{1}{7}\right)\) | M1 | \((=7)\) |
| \(y - 5.5 = 7(x-3.5)\) oe ISW | A1 | cao. Correct answer, no working or inadequate working: SC B2 |
| Midpt \(AB\) is \((3.5,5.5)\); Gradient \(AB=-\frac{1}{7}\) | B1 | Both |
| \((y=7x+c)\): \(5.5=7\times3.5+c\) | M1 | ft their midpt and gradient, NOT \(-\frac{1}{7}\) |
| \(y = 7x - 19\) | A1 | cao. Any correct form |
| \(x^2+(y-6)^2=(x-7)^2+(y-5)^2\) | M1, M1 | Attempt expansion |
| \(-12y+36=-14x-10y+49+25\) ISW | A1 | cao. Any correct form eg \(y=7x-19\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Perpendicular bisector of \(BC\) is \(x+7y-17=0\); OR of \(CA\) is \(4y=3x-1\) | B1 | Any correct form for another perp bisector |
| Example method, perp bisectors of \(AB\) & \(BC\): \(x+7(7x-19)-17=0\ (\Rightarrow x=3)\) | M1 | Attempt solve simultaneously equations of two perpendicular bisectors. Can be implied |
| Alternative for 1st two marks: Grad \(BC\) is 7 so \(BC\) & \(AB\) perpendicular. Hence \(AC\) is a diameter | M1, B1 | |
| Centre is \((3,2)\) | B1 | cao. NB if centre \(=(3,2)\) without clear working, B0M0B1 |
| eg Radius\(^2 = 3^2+(6-2)^2=25\) | M1 | Correct method for \(r^2\) or \(r\) using their centre & \(A\) or \(B\) or \(C\) |
| Equation of circle is \((x-3)^2+(y-2)^2=25\) or \(x^2-6x+y^2-4y=12\) oe | A1ft | ISW. ft their centre & radius, dep both M1 marks |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Midpoint $AB$ is $(3.5, 5.5)$; Gradient $AB = -\frac{1}{7}$ | **B1** | Both. Allow midpoint $= \left(\frac{0+7}{2}, \frac{6+5}{2}\right)$ ISW |
| Gradient of perpendicular bisector $-1/\left(-\frac{1}{7}\right)$ | **M1** | $(=7)$ |
| $y - 5.5 = 7(x-3.5)$ oe ISW | **A1** | cao. Correct answer, no working or inadequate working: SC B2 |
| Midpt $AB$ is $(3.5,5.5)$; Gradient $AB=-\frac{1}{7}$ | **B1** | Both |
| $(y=7x+c)$: $5.5=7\times3.5+c$ | **M1** | ft their midpt and gradient, NOT $-\frac{1}{7}$ |
| $y = 7x - 19$ | **A1** | cao. Any correct form |
| $x^2+(y-6)^2=(x-7)^2+(y-5)^2$ | **M1, M1** | Attempt expansion |
| $-12y+36=-14x-10y+49+25$ ISW | **A1** | cao. Any correct form eg $y=7x-19$ |
---
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Perpendicular bisector of $BC$ is $x+7y-17=0$; OR of $CA$ is $4y=3x-1$ | **B1** | Any correct form for another perp bisector |
| Example method, perp bisectors of $AB$ & $BC$: $x+7(7x-19)-17=0\ (\Rightarrow x=3)$ | **M1** | Attempt solve simultaneously equations of two perpendicular bisectors. Can be implied |
| **Alternative for 1st two marks:** Grad $BC$ is 7 so $BC$ & $AB$ perpendicular. Hence $AC$ is a diameter | **M1, B1** | |
| Centre is $(3,2)$ | **B1** | cao. NB if centre $=(3,2)$ without clear working, B0M0B1 |
| eg Radius$^2 = 3^2+(6-2)^2=25$ | **M1** | Correct method for $r^2$ or $r$ using their centre & $A$ or $B$ or $C$ |
| Equation of circle is $(x-3)^2+(y-2)^2=25$ or $x^2-6x+y^2-4y=12$ oe | **A1ft** | ISW. ft their centre & radius, dep both M1 marks |
---5 In this question you must show detailed reasoning.
Points $A , B$ and $C$ have coordinates $( 0,6 ) , ( 7,5 )$ and $( 6 , - 2 )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find an equation of the perpendicular bisector of $A B$.
\item Hence, or otherwise, find an equation of the circle that passes through points $A , B$ and $C$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2021 Q5 [8]}}