| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle through three points using perpendicular bisectors |
| Difficulty | Standard +0.3 This is a standard multi-part question on circle geometry requiring perpendicular bisector equations, distance formula, and circle equations. While it has several parts, each step follows routine procedures with no novel insight required. The perpendicular bisector work and showing equal distances are straightforward applications of coordinate geometry formulas, making it slightly easier than average. |
| 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^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks |
|---|---|
| \(P\) is \((3,2) \Rightarrow XP\) is \(x = 3 \Rightarrow X(3,3)\) | B1, B1, B1, M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(XA = \sqrt{(3-0)^2 + (3-2)^2} = \sqrt{10}\) | M1, A1 | One |
| \(XB = \sqrt{(3-2)^2 + (3-6)^2} = \sqrt{10}\) | A1 | Other two |
| Answer | Marks |
|---|---|
| Then probably in (iii), \(XA = \sqrt{(3-0)^2 + (3-2)^2} = \sqrt{10}\) | B1, M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((x - 3)^2 + (y - 3)^2 = 10\) | B1, B1 | LHS; RHS |
| Answer | Marks |
|---|---|
| \(\Rightarrow x = 4, 2\) | M1, A1 |
**Part (i)**
Grad $BC = -1 \Rightarrow$ Grad $XQ = 1$
$Q$ is $(4,4)$
$\Rightarrow y = x$
$P$ is $(3,2) \Rightarrow XP$ is $x = 3 \Rightarrow X(3,3)$ | B1, B1, B1, M1, A1 |
**Part (ii)**
$XA = \sqrt{(3-0)^2 + (3-2)^2} = \sqrt{10}$ | M1, A1 | One
$XB = \sqrt{(3-2)^2 + (3-6)^2} = \sqrt{10}$ | A1 | Other two
$XC = \sqrt{(3-6)^2 + (3-2)^2} = \sqrt{10}$
Alternative argument: $XA = XC$ because on $XP$; $XB = XC$ because on $XQ$; Therefore $XA = XB = XC$
Then probably in (iii), $XA = \sqrt{(3-0)^2 + (3-2)^2} = \sqrt{10}$ | B1, M1, A1 |
**Part (iii)**
$(x - 3)^2 + (y - 3)^2 = 10$ | B1, B1 | LHS; RHS
**Part (iv)**
$y = 0 \Rightarrow x^2 - 6x + 8 = 0$
$\Rightarrow x = 4, 2$ | M1, A1 |13 In Fig.13, XP and XQ are the perpendicular bisectors of AC and BC respectively.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4c556b8e-1a19-4480-bf2a-0ef9e67f98b4-5_409_768_383_604}
\captionsetup{labelformat=empty}
\caption{Fig. 13}
\end{center}
\end{figure}
(i) Find the coordinates of X .\\
(ii) Hence show that $\mathrm { AX } = \mathrm { BX } = \mathrm { CX }$.\\
(iii) The circumcircle of a triangle is the circle which passes through the vertices of the triangle.\\
Write down the equation of the circumcircle of the triangle ABC .\\
(iv) Find the coordinates of the points where the circle cuts the $x$ axis.
\hfill \mbox{\textit{OCR MEI C1 Q13 [12]}}