| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent from external point - intersection or geometric properties |
| Difficulty | Standard +0.3 This is a standard AS-level coordinate geometry question combining circle equations, tangent properties, and distance calculations. Part (a) requires simple radius verification (3 marks of routine calculation). Part (b) involves finding perpendicular gradients, tangent equations, and y-intercepts—all standard techniques with no novel insight required. The 10 total marks reflect length rather than conceptual difficulty, making this slightly easier than average. |
| Spec | 1.03b Straight lines: parallel and perpendicular relationships1.03d 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 equations1.10f Distance between points: using position vectors |
| Answer | Marks |
|---|---|
| 17 (a) | Way 1: |
| Answer | Marks |
|---|---|
| (10(cid:114)((cid:16)2))2 (cid:14)(11 6)2 | Way 2: |
| Answer | Marks | Guidance |
|---|---|---|
| ((cid:16)2, 6) and (10, 11) | M1 | 3.1a |
| Answer | Marks | Guidance |
|---|---|---|
| satisfies their circle equation | Finds distance between ((cid:16)2, 6) | |
| and (10, 1) | M1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| states that (10, 1) lies on C * | Concludes that as distance is the | |
| same (10, 1) lies on the circle C * | A1* | 2.1 |
| Answer | Marks |
|---|---|
| (b) | 11(cid:16)6 1(cid:16)6 |
| Answer | Marks | Guidance |
|---|---|---|
| 10(cid:16)((cid:16)2) 10(cid:16)((cid:16)2) | M1 | 3.1a |
| Answer | Marks | Guidance |
|---|---|---|
| m | M1 | 1.1b |
| Finds (equation and ) y intercept of tangent (see note below) | M1 | 1.1b |
| Obtains a correct value for y intercept of their tangent i.e.35 or –23 | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| second tangent | Way 2: Deduces midpoint of PQ | |
| from symmetry (0, 6) | M1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| intercept of second tangent | Uses this to find other intercept | M1 |
| So obtains distance PQ = 35 + 23= 58* | A1* | 1.1b |
Question 17:
--- 17 (a) ---
17 (a) | Way 1:
Finds circle equation
(x(cid:114)2)2 (cid:14)(y 6)2 (cid:32)
(10(cid:114)((cid:16)2))2 (cid:14)(11 6)2 | Way 2:
Finds distance between
((cid:16)2, 6) and (10, 11) | M1 | 3.1a
Checks whether (10, 1)
satisfies their circle equation | Finds distance between ((cid:16)2, 6)
and (10, 1) | M1 | 1.1b
Obtains
(x(cid:14)2)2 (cid:14)(y(cid:16)6)2 (cid:32)132
and checks that
(10(cid:14)2)2 (cid:14)(1(cid:16)6)2 (cid:32)132so
states that (10, 1) lies on C * | Concludes that as distance is the
same (10, 1) lies on the circle C * | A1* | 2.1
(3)
(b) | 11(cid:16)6 1(cid:16)6
Finds radius gradient or (m)
10(cid:16)((cid:16)2) 10(cid:16)((cid:16)2) | M1 | 3.1a
1
Finds gradient perpendicular to their radius using (cid:16)
m | M1 | 1.1b
Finds (equation and ) y intercept of tangent (see note below) | M1 | 1.1b
Obtains a correct value for y intercept of their tangent i.e.35 or –23 | A1 | 1.1b
Way 1: Deduces gradient of
second tangent | Way 2: Deduces midpoint of PQ
from symmetry (0, 6) | M1 | 1.1b
Finds (equation and ) y
intercept of second tangent | Uses this to find other intercept | M1 | 1.1b
So obtains distance PQ = 35 + 23= 58* | A1* | 1.1b
(7)
(10 marks)
Question 17 continued
Notes:
(a) Way 1 and Way 2:
M1: Starts to use information in question to find equation of circle or radius of circle
M1: Completes method for checking that (10, 1) lies on circle
A1*: Completely correct explanation with no errors concluding with statement that circle
passes through (10, 1)
(b)
11(cid:16)6 1(cid:16)6
M1: Calculates or (m)
10(cid:16)((cid:16)2) 10(cid:16)((cid:16)2)
1 12 12
M1: Finds (cid:16) (correct answer is (cid:16) or ). This is referred to as m(cid:99)in the next note
m 5 5
(cid:167) 12(cid:183) (cid:167)12(cid:183)
M1: Attempts y(cid:16)11(cid:32)their (cid:168) (cid:16) (cid:184) (x(cid:16)10) or y(cid:16)1(cid:32)their (cid:168) (cid:184) (x(cid:16)10)and puts x = 0, or
(cid:169) 5 (cid:185) (cid:169) 5 (cid:185)
y(cid:16)11
uses vectors to find intercept e.g. (cid:32)(cid:16)m(cid:99)
10
A1: One correct intercept 35 or – 23
Way 1:
12 12
M1: Uses the negative of their previous tangent gradient or uses a correct (cid:16) or
5 5
M1: Attempts the second tangent equation and puts x = 0 or uses vectors to find intercept
11(cid:16) y
e.g. (cid:32)m(cid:99)
10
Way 2:
M1: Finds midpoint of PQ from symmetry. (This is at (0, 6))
M1: Uses this midpoint to find second intercept or to find difference between midpoint and
first intercept. e.g. 35 – 6 = 29 then 6 – 29 = –23 so second intercept is at (–23, 0)
Ways 1 and 2:
A1*: Obtain 58 correctly from a valid methodA circle $C$ with centre at $(-2, 6)$ passes through the point $(10, 11)$.
\begin{enumerate}[label=(\alph*)]
\item Show that the circle $C$ also passes through the point $(10, 1)$.
[3]
\end{enumerate}
The tangent to the circle $C$ at the point $(10, 11)$ meets the $y$ axis at the point $P$ and the tangent to the circle $C$ at the point $(10, 1)$ meets the $y$ axis at the point $Q$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the distance $PQ$ is $58$ explaining your method clearly.
[7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 Q17 [10]}}