| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| 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 straightforward multi-part circle question testing standard C1 techniques: expanding circle equation, substituting to find coordinates, distance comparison, and tangent using perpendicular gradient. All parts are routine applications with no problem-solving insight required, making it slightly easier than average but not trivial due to the algebraic manipulation needed in parts (ii) and (iv). |
| 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/Working | Mark | Guidance |
| \((x-2)^2\) and \((y-1)^2\) seen | B1 | |
| \((x-2)^2 + (y-1)^2 = 100\) | B1 | \((x \pm 2)^2 + (y \pm 1)^2 = 100\) |
| \(x^2 + y^2 - 4x - 2y - 95 = 0\) | B1 | Correct form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((5-2)^2 + (k-1)^2 = 100\) | M1 | \(x = 5\) substituted into their equation |
| \((k-1)^2 = 91\) or \(k^2 - 2k - 90 = 0\) | A1 | Correct simplified quadratic in \(k\) (or \(y\)) obtained |
| \(k = 1 + \sqrt{91}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance from \((-3, 9)\) to \((2, 1)\): \(= \sqrt{(2--3)^2 + (1-9)^2}\) | M1 | Uses \((x_2 - x_1)^2 + (y_2 - y_1)^2\) |
| \(= \sqrt{25 + 64} = \sqrt{89}\) | A1 | |
| \(\sqrt{89} < 10\) so point is inside | B1 | Compares their distance with 10 and makes consistent conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gradient of radius \(= \frac{9-1}{8-2}\) | M1 | Uses \(\frac{y_2 - y_1}{x_2 - x_1}\) |
| \(= \frac{4}{3}\) | A1 | oe |
| Gradient of tangent \(= -\frac{3}{4}\) | B1\(\checkmark\) | oe |
| \(y - 9 = -\frac{3}{4}(x - 8)\) | M1 | Correct equation of straight line through \((8, 9)\), any non-zero gradient |
| \(y = -\frac{3}{4}x + 15\) | A1 | oe 3 term equation |
## Question 9(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x-2)^2$ and $(y-1)^2$ seen | B1 | |
| $(x-2)^2 + (y-1)^2 = 100$ | B1 | $(x \pm 2)^2 + (y \pm 1)^2 = 100$ |
| $x^2 + y^2 - 4x - 2y - 95 = 0$ | B1 | Correct form |
---
## Question 9(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(5-2)^2 + (k-1)^2 = 100$ | M1 | $x = 5$ substituted into their equation |
| $(k-1)^2 = 91$ or $k^2 - 2k - 90 = 0$ | A1 | Correct simplified quadratic in $k$ (or $y$) obtained |
| $k = 1 + \sqrt{91}$ | A1 | cao |
---
## Question 9(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance from $(-3, 9)$ to $(2, 1)$: $= \sqrt{(2--3)^2 + (1-9)^2}$ | M1 | Uses $(x_2 - x_1)^2 + (y_2 - y_1)^2$ |
| $= \sqrt{25 + 64} = \sqrt{89}$ | A1 | |
| $\sqrt{89} < 10$ so point is inside | B1 | Compares their distance with 10 and makes consistent conclusion |
---
## Question 9(iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of radius $= \frac{9-1}{8-2}$ | M1 | Uses $\frac{y_2 - y_1}{x_2 - x_1}$ |
| $= \frac{4}{3}$ | A1 | oe |
| Gradient of tangent $= -\frac{3}{4}$ | B1$\checkmark$ | oe |
| $y - 9 = -\frac{3}{4}(x - 8)$ | M1 | Correct equation of straight line through $(8, 9)$, any non-zero gradient |
| $y = -\frac{3}{4}x + 15$ | A1 | oe 3 term equation |
---9 (i) Find the equation of the circle with radius 10 and centre ( 2,1 ), giving your answer in the form $x ^ { 2 } + y ^ { 2 } + a x + b y + c = 0$.\\
(ii) The circle passes through the point $( 5 , k )$ where $k > 0$. Find the value of $k$ in the form $p + \sqrt { q }$.\\
(iii) Determine, showing all working, whether the point $( - 3,9 )$ lies inside or outside the circle.\\
(iv) Find an equation of the tangent to the circle at the point ( 8,9 ).
\hfill \mbox{\textit{OCR C1 2008 Q9 [14]}}