| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Moderate -0.8 This is a straightforward multi-part circle question testing basic concepts: reading center/radius from standard form (trivial), finding intersections by substitution (routine), checking if a point lies on/inside/outside (simple calculation), and finding a tangent equation using perpendicular gradient (standard technique). All parts are textbook exercises requiring only direct application of formulas with no problem-solving or insight needed. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Radius \(= 5\) | B1 | |
| \((2, -3)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((y+3)^2 = 21\) or \(y^2 + 6y - 12 = 0\) | M1 | Substituting \(x = 0\) and rearranging |
| Roots \(-3 + \sqrt{21}\) and \(-3 - \sqrt{21}\) | A1 | For one \(y\)-value |
| \(\left(0, -3+\sqrt{21}\right)\) and \(\left(0, -3-\sqrt{21}\right)\) | A1 | All correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((1-2)^2 + (2+3)^2 = 1^2 + 5^2\) | M1 | Or distance of \((1,2)\) from their centre |
| This is more than \(25\) so outside the circle | A1 | Or distance of \((1,2)\) from centre is \(\sqrt{26} > 5\), so outside the circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Gradient \(CP = \frac{1-(-3)}{-1-2} = -\frac{4}{3}\) FT their centre | M1 | |
| Gradient of tangent \(= \frac{3}{4}\) FT their grad CP | M1 | |
| Equation of tangent \(y - 1 = \frac{3}{4}(x-(-1))\) FT their grad | M1 | |
| \(4y = 3x + 7\) oe | A1 |
# Question 8:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Radius $= 5$ | B1 | |
| $(2, -3)$ | B1 | |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(y+3)^2 = 21$ or $y^2 + 6y - 12 = 0$ | M1 | Substituting $x = 0$ and rearranging |
| Roots $-3 + \sqrt{21}$ and $-3 - \sqrt{21}$ | A1 | For one $y$-value |
| $\left(0, -3+\sqrt{21}\right)$ and $\left(0, -3-\sqrt{21}\right)$ | A1 | All correct |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1-2)^2 + (2+3)^2 = 1^2 + 5^2$ | M1 | Or distance of $(1,2)$ from their centre |
| This is more than $25$ so outside the circle | A1 | Or distance of $(1,2)$ from centre is $\sqrt{26} > 5$, so outside the circle |
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient $CP = \frac{1-(-3)}{-1-2} = -\frac{4}{3}$ FT their centre | M1 | |
| Gradient of tangent $= \frac{3}{4}$ FT their grad CP | M1 | |
| Equation of tangent $y - 1 = \frac{3}{4}(x-(-1))$ FT their grad | M1 | |
| $4y = 3x + 7$ oe | A1 | |
---8 A circle has equation $( x - 2 ) ^ { 2 } + ( y + 3 ) ^ { 2 } = 25$.
\begin{enumerate}[label=(\alph*)]
\item Write down
\begin{itemize}
\item the radius of the circle,
\item the coordinates of the centre of the circle.
\item Find, in exact form, the coordinates of the points of intersection of the circle with the $y$-axis.
\item Show that the point $( 1,2 )$ lies outside the circle.
\item The point $\mathrm { P } ( - 1,1 )$ lies on the circle. Find the equation of the tangent to the circle at P .
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 Q8 [11]}}