| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Chord length calculation |
| Difficulty | Standard +0.3 This is a straightforward multi-part circle question requiring standard techniques: reading centre/radius from equation form, finding axis intersections by substitution, verifying points lie on the circle, and using the perpendicular-from-centre-to-chord property. All steps are routine C1 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| [radius \(=\)] \(\sqrt{20}\) or \(2\sqrt{5}\) isw | B1 | B0 for \(\pm\sqrt{20}\) oe |
| [centre \(=\)] \((3, 2)\) | B1 | condone lack of brackets with coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| substitution of \(x = 0\) or \(y = 0\) into circle equation | M1 | or use of Pythagoras with radius and a coordinate of the centre eg \(20 - 2^2\) or \(h^2 + 3^2 = 20\) ft their centre and/or radius; equation may be expanded first; allow M1 for \((x-3)^2 = 20\) and/or \((y-2)^2 = 20\) |
| \((x-7)(x+1) [=0]\) | M1 | no ft from wrong quadratic; for factors giving two terms correct, or formula or completing square used with at most one error; completing square attempt must reach at least \((x-a)^2 = b\) |
| \((7, 0)\) and \((-1, 0)\) isw | A1 | accept \(x = 7\) or \(-1\) (both required) |
| \([y =] \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times (-7)}}{2}\) oe | M1 | no ft from wrong quadratic; for formula or completing square used with at most one error; completing square attempt must reach at least \((y-a)^2 = b\); following use of Pythagoras allow M1 for attempt to add 2 to \([\pm]\sqrt{11}\) |
| \(\left(0, 2 \pm \sqrt{11}\right)\) or \(\left(0, \frac{4 \pm \sqrt{44}}{2}\right)\) isw | A1 | accept \(y = \frac{4 \pm \sqrt{44}}{2}\) oe isw; annotation required if part marks earned |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| show both A and B are on circle | B1 | explicit substitution in circle equation and at least one stage of interim working required oe; or clear use of Pythagoras to show AC and BC each \(= \sqrt{20}\) |
| \((4, 5)\) | B2 | B1 each; or M1 for \(\left(\frac{7+1}{2}, \frac{6+4}{2}\right)\) |
| \(\sqrt{10}\) | B2 | from correct midpoint and centre used; B1 for \(\pm\sqrt{10}\); M1 for \((4-3)^2 + (5-2)^2\) or \(1^2 + 3^2\) or ft their centre and/or midpoint, or for the square root of this; no ft if one coord of midpoint is same as that of centre so that distance formula/Pythag is not required eg centre correct and midpt \((3, -1)\); annotation required if part marks earned |
## Question 3(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| [radius $=$] $\sqrt{20}$ or $2\sqrt{5}$ isw | B1 | B0 for $\pm\sqrt{20}$ oe |
| [centre $=$] $(3, 2)$ | B1 | condone lack of brackets with coordinates |
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## Question 3(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| substitution of $x = 0$ or $y = 0$ into circle equation | M1 | or use of Pythagoras with radius and a coordinate of the centre eg $20 - 2^2$ or $h^2 + 3^2 = 20$ ft their centre and/or radius; equation may be expanded first; allow M1 for $(x-3)^2 = 20$ and/or $(y-2)^2 = 20$ |
| $(x-7)(x+1) [=0]$ | M1 | no ft from wrong quadratic; for factors giving two terms correct, or formula or completing square used with at most one error; completing square attempt must reach at least $(x-a)^2 = b$ |
| $(7, 0)$ and $(-1, 0)$ isw | A1 | accept $x = 7$ or $-1$ (both required) |
| $[y =] \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times (-7)}}{2}$ oe | M1 | no ft from wrong quadratic; for formula or completing square used with at most one error; completing square attempt must reach at least $(y-a)^2 = b$; following use of Pythagoras allow M1 for attempt to add 2 to $[\pm]\sqrt{11}$ |
| $\left(0, 2 \pm \sqrt{11}\right)$ or $\left(0, \frac{4 \pm \sqrt{44}}{2}\right)$ isw | A1 | accept $y = \frac{4 \pm \sqrt{44}}{2}$ oe isw; annotation required if part marks earned |
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## Question 3(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| show both A and B are on circle | B1 | explicit substitution in circle equation and at least one stage of interim working required oe; or clear use of Pythagoras to show AC and BC each $= \sqrt{20}$ |
| $(4, 5)$ | B2 | B1 each; or M1 for $\left(\frac{7+1}{2}, \frac{6+4}{2}\right)$ |
| $\sqrt{10}$ | B2 | from correct midpoint and centre used; B1 for $\pm\sqrt{10}$; M1 for $(4-3)^2 + (5-2)^2$ or $1^2 + 3^2$ or ft their centre and/or midpoint, or for the square root of this; no ft if one coord of midpoint is same as that of centre so that distance formula/Pythag is not required eg centre correct and midpt $(3, -1)$; annotation required if part marks earned |3 The circle $( x - 3 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 20$ has centre C.\\
(i) Write down the radius of the circle and the coordinates of C .\\
(ii) Find the coordinates of the intersections of the circle with the $x$ - and $y$-axes.\\
(iii) Show that the points $\mathrm { A } ( 1,6 )$ and $\mathrm { B } ( 7,4 )$ lie on the circle. Find the coordinates of the midpoint of AB . Find also the distance of the chord AB from the centre of the circle.
\hfill \mbox{\textit{OCR MEI C1 Q3 [12]}}