| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Moderate -0.8 This is a straightforward C1 question testing basic coordinate geometry: substituting into a linear equation, distance formula, reading circle equation in completed square form, and verifying a diameter by checking the centre is the midpoint. All parts are routine recall and calculation with no problem-solving required, making it easier than average but not trivial since it requires multiple standard techniques. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(30+4k-10=0\) | M1 | Attempt to substitute \(x=10\) into equation of line |
| \(\therefore k=-5\) | A1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sqrt{(10-2)^2+(-5-1)^2}\) | M1 | Correct method to find line length using Pythagoras' theorem |
| \(=\sqrt{64+36}=10\) | A1 (2) | cao, dependent on correct value of k in (i) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Centre \((6,-2)\) | B1 | |
| Radius \(5\) | B1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Midpoint of \(AB = (6,-2)\) | B1 | One correct statement of verification |
| Length of \(AB = 2\times\) radius; Both A and B lie on circumference; Centre lies on line \(3x+4y-10=0\) | B1 (2) | Complete verification |
## Question 7:
**Part (i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $30+4k-10=0$ | M1 | Attempt to substitute $x=10$ into equation of line |
| $\therefore k=-5$ | A1 (2) | |
**Part (ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sqrt{(10-2)^2+(-5-1)^2}$ | M1 | Correct method to find line length using Pythagoras' theorem |
| $=\sqrt{64+36}=10$ | A1 (2) | cao, dependent on correct value of k in (i) |
**Part (iii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Centre $(6,-2)$ | B1 | |
| Radius $5$ | B1 (2) | |
**Part (iv):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Midpoint of $AB = (6,-2)$ | B1 | One correct statement of verification |
| Length of $AB = 2\times$ radius; Both A and B lie on circumference; Centre lies on line $3x+4y-10=0$ | B1 (2) | Complete verification |
---7 The line with equation $3 x + 4 y - 10 = 0$ passes through point $A ( 2,1 )$ and point $B ( 10 , k )$.\\
(i) Find the value of $k$.\\
(ii) Calculate the length of $A B$.
A circle has equation $( x - 6 ) ^ { 2 } + ( y + 2 ) ^ { 2 } = 25$.\\
(iii) Write down the coordinates of the centre and the radius of the circle.\\
(iv) Verify that $A B$ is a diameter of the circle.
\hfill \mbox{\textit{OCR C1 2009 Q7 [8]}}