| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Moderate -0.3 This is a multi-part circle question requiring completing the square, finding tangent equations using perpendicular gradients, and geometric reasoning about concentric circles. While it has multiple parts (typical of 8-10 mark questions), each step uses standard C1 techniques without requiring novel insight. The tangent work is routine (radius perpendicular to tangent), and part (iv) requires basic geometric understanding of distance between centers. Slightly easier than average due to straightforward application of standard methods. |
| 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, tangents1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Centre of circle \((4, 3)\) | B1 | Correct centre |
| \((x-4)^2 - 16 + (y-3)^2 - 9 - 20 = 0\) | M1 | \((x\pm4)^2 - 4^2\) and \((y\pm3)^2 - 3^2\) seen (or implied by correct answer). Or \(r^2 = 4^2 + 3^2 + 20\) soi |
| \(r^2 = 45\), \(r = \sqrt{45}\) | A1 | \(\sqrt{45}\) or better www. ISW after \(\sqrt{45}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At A, \(y=0\) so \(x^2 - 8x - 20 = 0\); \((x-10)(x+2)=0\) | M1 | Valid method to find A e.g. put \(y=0\) and attempt to solve quadratic (allow slips) or Pythagoras' theorem |
| \(A = (10, 0)\) | A1 | Correct answer found |
| Gradient of radius \(= \frac{3-0}{4-10} = -\frac{1}{2}\) | M1 | Attempts to find gradient of radius (3 out of 4 terms correct for their centre, their A). Alternative: M1 attempt at implicit differentiation as evidenced by \(2y\frac{dy}{dx}\) term; A1 \(2x + 2y\frac{dy}{dx} - 8 - 6\frac{dy}{dx} = 0\) and substitution of \((10,0)\) to obtain 2. |
| Gradient of tangent \(= 2\) | B1 | |
| \(y - 0 = 2(x-10)\); \(y = 2x - 20\) | M1, A1 | Equation of line through their A, any non-zero gradient. Correct answer in any three-term form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A' = (-2, 6)\) | B1 | Finds the opposite end of the diameter |
| \(y - 6 = 2(x+2)\); \(y = 2x + 10\) | M1, A1 | Line through their A' parallel to their line in (ii). Correct answer in any three-term form. Not through centre of circle. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(OC = \sqrt{3^2 + 4^2} = 5\) | M1 | Attempts to find distance from O to their centre and subtract from their radius |
| \((0 <)\ r < \sqrt{45} - 5\) | A1 | Correct inequality, condone \(\leq\). ISW incorrect simplification |
## Question 10:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Centre of circle $(4, 3)$ | B1 | Correct centre |
| $(x-4)^2 - 16 + (y-3)^2 - 9 - 20 = 0$ | M1 | $(x\pm4)^2 - 4^2$ and $(y\pm3)^2 - 3^2$ seen (or implied by correct answer). Or $r^2 = 4^2 + 3^2 + 20$ soi |
| $r^2 = 45$, $r = \sqrt{45}$ | A1 | $\sqrt{45}$ or better www. ISW after $\sqrt{45}$ |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At A, $y=0$ so $x^2 - 8x - 20 = 0$; $(x-10)(x+2)=0$ | M1 | Valid method to find A e.g. put $y=0$ and attempt to solve quadratic (allow slips) or Pythagoras' theorem |
| $A = (10, 0)$ | A1 | Correct answer found |
| Gradient of radius $= \frac{3-0}{4-10} = -\frac{1}{2}$ | M1 | Attempts to find gradient of radius (3 out of 4 terms correct for their centre, their A). **Alternative:** M1 attempt at implicit differentiation as evidenced by $2y\frac{dy}{dx}$ term; A1 $2x + 2y\frac{dy}{dx} - 8 - 6\frac{dy}{dx} = 0$ and substitution of $(10,0)$ to obtain 2. |
| Gradient of tangent $= 2$ | B1 | |
| $y - 0 = 2(x-10)$; $y = 2x - 20$ | M1, A1 | Equation of line through their A, any non-zero gradient. Correct answer in any three-term form |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A' = (-2, 6)$ | B1 | Finds the opposite end of the diameter |
| $y - 6 = 2(x+2)$; $y = 2x + 10$ | M1, A1 | Line through their A' parallel to their line in (ii). Correct answer in any three-term form. Not through centre of circle. |
### Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $OC = \sqrt{3^2 + 4^2} = 5$ | M1 | Attempts to find distance from O to their centre **and subtract from their radius** |
| $(0 <)\ r < \sqrt{45} - 5$ | A1 | Correct inequality, condone $\leq$. ISW incorrect simplification |10\\
\includegraphics[max width=\textwidth, alt={}, center]{0ae3af7e-32cc-43fa-89bb-d6697a8f8061-3_755_905_248_580}
The diagram shows the circle with equation $x ^ { 2 } + y ^ { 2 } - 8 x - 6 y - 20 = 0$.\\
(i) Find the centre and radius of the circle.
The circle crosses the positive $x$-axis at the point $A$.\\
(ii) Find the equation of the tangent to the circle at $A$.\\
(iii) A second tangent to the circle is parallel to the tangent at $A$. Find the equation of this second tangent.\\
(iv) Another circle has centre at the origin $O$ and radius $r$. This circle lies wholly inside the first circle. Find the set of possible values of $r$.
\hfill \mbox{\textit{OCR C1 2016 Q10 [14]}}