| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Line-circle intersection points |
| Difficulty | Moderate -0.8 This is a straightforward C1 question requiring basic circle recognition and solving simultaneous equations by substitution. The circle equation is in standard form, the line equation is simple, and the algebraic manipulation (substituting y = 5 - 2x into the circle equation) leads to a routine quadratic. Below average difficulty as it's a standard textbook exercise with no conceptual challenges. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02p Interpret algebraic solutions: graphically1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Circle, centre \((0, 0)\), radius \(5\) | B1 | Circle centre \((0,0)\) |
| B1 2 | Radius 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = 5 - 2x\) | M1 | Attempt to solve equations simultaneously |
| \(x^2 + (5-2x)^2 = 25\) | ||
| \(5x^2 - 20x = 0\) | *M1 | Substitute for \(x/y\) or correct attempt at elimination of one variable (NOT for 2 linear equations) |
| OR: \(x = \frac{5-y}{2}\), \(\frac{(5-y)^2}{4} + y^2 = 25\) | DM1 | Obtain quadratic \(ax^2 + bx + c = 0\) (\(a \neq 0, b \neq 0\)) |
| \(y^2 - 2y - 15 = 0\) | M1 | Correct method to solve quadratic |
| \(x = 0, 4\) | A1 | \(x = 0, 4\) or \(y = 5, -3\) |
| \(y = 5, -3\) | A1 6 | \(y = 5, -3\) or \(x = 0, 4\). SR: one correct pair www B1 |
## Question 8(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Circle, centre $(0, 0)$, radius $5$ | B1 | Circle centre $(0,0)$ |
| | B1 **2** | Radius 5 |
## Question 8(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = 5 - 2x$ | M1 | Attempt to solve equations simultaneously |
| $x^2 + (5-2x)^2 = 25$ | | |
| $5x^2 - 20x = 0$ | *M1 | Substitute for $x/y$ or correct attempt at elimination of one variable (NOT for 2 linear equations) |
| **OR:** $x = \frac{5-y}{2}$, $\frac{(5-y)^2}{4} + y^2 = 25$ | DM1 | Obtain quadratic $ax^2 + bx + c = 0$ ($a \neq 0, b \neq 0$) |
| $y^2 - 2y - 15 = 0$ | M1 | Correct method to solve quadratic |
| $x = 0, 4$ | A1 | $x = 0, 4$ or $y = 5, -3$ |
| $y = 5, -3$ | A1 **6** | $y = 5, -3$ or $x = 0, 4$. SR: one correct pair www **B1** |
---8 (i) Describe completely the curve $x ^ { 2 } + y ^ { 2 } = 25$.\\
(ii) Find the coordinates of the points of intersection of the curve $x ^ { 2 } + y ^ { 2 } = 25$ and the line $2 x + y - 5 = 0$.
\hfill \mbox{\textit{OCR C1 2005 Q8 [8]}}