| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Moderate -0.3 This is a straightforward multi-part circle question requiring standard techniques: midpoint/distance formula for center and radius, expanding circle equation, and finding a tangent line using perpendicular gradients. All methods are routine C1 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(AB^2 = (10-2)^2 + (3-9)^2 = 100\). Hence the radius is \(5\). Mid-point of \(AB\) is \(\left(\frac{2+10}{2}, \frac{9+3}{2}\right)\). Hence centre is \((6, 6)\) | M1 A1 M1 A1 | For correct calculation method for \(AB^2\); For correct value for radius; For correct calculation method for mid-point; For both coordinates correct |
| 4 | ||
| (ii) Equation is \((x-6)^2 + (y-6)^2 = 5^2\). This is \(x^2 - 12x + 36 + y^2 - 12y + 36 = 25\), i.e. \(x^2 + y^2 - 12x - 12y + 47 = 0\), as required | M1 A1 A1 | For using correct basic form of circle eqn; For expanding at least one bracket correctly; For showing given answer correctly |
| 3 | ||
| (iii) Gradient of \(AB\) is \(\frac{3-9}{10-2} = -\frac{3}{4}\). Hence perpendicular gradient is \(\frac{4}{3}\). Equation of tangent is \(y - 3 = \frac{4}{3}(x-10)\). Hence \(C\) is the point \(\left(\frac{41}{4}, 0\right)\) | M1 A1 A1 √ M1 M1 A1 | For finding the gradient of \(AB\); For correct value \(-\frac{3}{4}\) or equivalent; For relevant perpendicular gradient; For using their perp grad and \(B\) correctly; For substituting \(y = 0\) in their tangent eqn; For correct value \(x = \frac{41}{4}\) |
| 6 |
(i) $AB^2 = (10-2)^2 + (3-9)^2 = 100$. Hence the radius is $5$. Mid-point of $AB$ is $\left(\frac{2+10}{2}, \frac{9+3}{2}\right)$. Hence centre is $(6, 6)$ | M1 A1 M1 A1 | For correct calculation method for $AB^2$; For correct value for radius; For correct calculation method for mid-point; For both coordinates correct
| | 4 |
(ii) Equation is $(x-6)^2 + (y-6)^2 = 5^2$. This is $x^2 - 12x + 36 + y^2 - 12y + 36 = 25$, i.e. $x^2 + y^2 - 12x - 12y + 47 = 0$, as required | M1 A1 A1 | For using correct basic form of circle eqn; For expanding at least one bracket correctly; For showing given answer correctly
| | 3 |
(iii) Gradient of $AB$ is $\frac{3-9}{10-2} = -\frac{3}{4}$. Hence perpendicular gradient is $\frac{4}{3}$. Equation of tangent is $y - 3 = \frac{4}{3}(x-10)$. Hence $C$ is the point $\left(\frac{41}{4}, 0\right)$ | M1 A1 A1 √ M1 M1 A1 | For finding the gradient of $AB$; For correct value $-\frac{3}{4}$ or equivalent; For relevant perpendicular gradient; For using their perp grad and $B$ correctly; For substituting $y = 0$ in their tangent eqn; For correct value $x = \frac{41}{4}$
| | 6 |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{5fa27228-37b2-45d9-a8dc-355b2f7f6fa4-3_757_810_1050_680}
The diagram shows a circle which passes through the points $A ( 2,9 )$ and $B ( 10,3 ) . A B$ is a diameter of the circle.\\
(i) Calculate the radius of the circle and the coordinates of the centre.\\
(ii) Show that the equation of the circle may be written in the form $x ^ { 2 } + y ^ { 2 } - 12 x - 12 y + 47 = 0$.\\
(iii) The tangent to the circle at the point $B$ cuts the $x$-axis at $C$. Find the coordinates of $C$.
\hfill \mbox{\textit{OCR C1 Q7 [13]}}