| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent from external point - find equation |
| Difficulty | Standard +0.3 This is a straightforward C1 circle question requiring: (i) substitution to find k, (ii) completing the square to find centre and radius (standard technique), and (iii) using Pythagoras on the right-angled triangle formed by centre-tangent point-external point. All are routine applications of standard methods with no novel insight 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 circle1.03f Circle properties: angles, chords, tangents |
| Answer | Marks |
|---|---|
| \((-6, 5)\) ∴ \(36 + 25 - 60 - 40 + k = 0\), \(k = 39\) | M1 A1 |
| Answer | Marks |
|---|---|
| \((x + 5)^2 - 25 + (y - 4)^2 - 16 + 39 = 0\) | M1 |
| \((x + 5)^2 + (y - 4)^2 = 2\) | M1 A1 |
| ∴ centre \((-5, 4)\), radius \(= \sqrt{2}\) | A2 |
| Answer | Marks |
|---|---|
| dist. \((2, 3)\) to centre \(= \sqrt{49 + 1} = \sqrt{50}\) | B1 |
| ∴ \(AB^2 = (\sqrt{50})^2 - (\sqrt{2})^2 = 48\) | M1 A1 |
| \(AB = \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}\) | M1 A1 |
| (10) |
## (i)
$(-6, 5)$ ∴ $36 + 25 - 60 - 40 + k = 0$, $k = 39$ | M1 A1
## (ii)
$(x + 5)^2 - 25 + (y - 4)^2 - 16 + 39 = 0$ | M1
$(x + 5)^2 + (y - 4)^2 = 2$ | M1 A1
∴ centre $(-5, 4)$, radius $= \sqrt{2}$ | A2
## (iii)
dist. $(2, 3)$ to centre $= \sqrt{49 + 1} = \sqrt{50}$ | B1
∴ $AB^2 = (\sqrt{50})^2 - (\sqrt{2})^2 = 48$ | M1 A1
$AB = \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$ | M1 A1
| (10)
---The circle $C$ has the equation
$$x^2 + y^2 + 10x - 8y + k = 0,$$
where $k$ is a constant.
Given that the point with coordinates $(-6, 5)$ lies on $C$,
\begin{enumerate}[label=(\roman*)]
\item find the value of $k$, [2]
\item find the coordinates of the centre and the radius of $C$. [3]
\end{enumerate}
A straight line which passes through the point $A(2, 3)$ is a tangent to $C$ at the point $B$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the length $AB$ in the form $k\sqrt{5}$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 Q8 [10]}}