EdExcel Core 2
Geometry and trigonometry
Topic assessment
1. Describe fully the curve whose equation is x 2  y 2  4 .
[2]
2. Show that the line y = 3x – 10 is a tangent to the circle x 2  y 2  10 .
[4]
3. The equation of a circle is x 2  y 2  4 x  2 y  15
(i) Find the coordinates of the centre C of the circle, and the radius of the circle.
(ii) Show that the point P (4, -5) lies on the circle.
(iii) Find the equation of the tangent to the circle at the point P.
[3]
[1]
[4]
4. AB is the diameter of a circle. A is (1, 3) and B is (7, -1).
(i) Find the coordinates of the centre C of the circle.
[2]
(ii) Find the radius of the circle.
[2]
(iii) Find the equation of the circle.
[2]
(iv) The line y + 5x = 8 cuts the circle at A and again at a second point D. Calculate the
coordinates of D.
[4]
(v) Prove that the line AB is perpendicular to the line CD.
[3]
5. Find the angle  and the length x in the triangle shown below.
B
52°
12 cm
x cm
A
15 cm

C
[7]
6. Andrew walks 5 km on a bearing of 140°, and then walks 3 km on a bearing of 025°.
(i) How far is Andrew from his starting point?
[3]
(ii) On what bearing should Andrew walk to get back to his starting point?
[4]
7. A belt is wrapped around a cylinder of radius 2.5 m as shown.
5m
Find the length of the belt.
[5]
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EdExcel C2 Geometry & trig Assessment solutions
8. Find the perimeter and area of the shaded sections of these shapes.
(i)
3 cm
60°
[7]
(ii)
8 cm
45°
10 cm
[7]
Total 60 marks
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EdExcel C2 Geometry & trig Assessment solutions
Geometry and trigonometry
Solutions to Topic assessment
1. The curve is a circle, centre O and radius 2.
[2]
2. Substituting y  3 x  10 into x 2  y 2  10
gives x 2  (3 x  10)2  10
x 2  9 x 2  60 x  100  10
10 x 2  60 x  90  0
x 2  6x  9  0
( x  3)2  0
Since the equation has a repeated root, the line meets the circle just once, and so the
line is a tangent to the circle.
[4]
3. (i) x 2  y 2  4 x  2 y  15
x 2  4 x  y 2  2 y  15
( x  2)2  4  ( y  1)2  1  15
( x  2)2  ( y  1)2  20
The centre C of the circle is (2, -1) and the radius is
20 .
[3]
(ii) Substituting x = 4 and y = -5: (4  2)  ( 5  1)  4  16  20
so the point (4, -5) lies on the circle.
2
2
[1]
1  ( 5 ) 4

 2
24
2
Tangent at P is perpendicular to CP, so gradient of tangent  21 .
Equation of tangent is y  ( 5 )  21 ( x  4)
(iii)Gradient of CP 
2( y  5 )  x  4
2 y  10  x  4
2 y  x  14
[4]
4. (i) C is the midpoint of AB.
1  7 3  ( 1) 
C  
,
  (4, 1)
2
 2

[2]
(ii) Radius of circle = CA  (4  1)  (1  3) 
2
2
9  4  13
(iii) Equation of circle is ( x  4)  ( y  1)  13
[2]
[2]
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EdExcel C2 Geometry & trig Assessment solutions
(iv) Substituting y  5 x  8 into equation of circle:
( x  4)2  ( 5 x  8  1)2  13
( x  4)2  ( 5 x  7)2  13
x 2  8 x  16  25 x 2  70 x  49  13
26 x 2  78 x  52  0
x 2  3x  2  0
( x  1)( x  2)  0
x  1 or x  2
x = 1 is point A, so point D is x = 2
When x = 2, y  5  2  8  2
The coordinates of D are (2, -2)
3 ( 1) 4
2


17
6
3
1  ( 2) 3

Gradient of CD 
42
2
[4]
(v) Gradient of AB 
Gradient of AB × gradient of CD  
so AB is perpendicular to CD.
2 3
  1
3 2
[3]
5. Using the sine rule:
sin  sin 52

12
15
12 sin 52
sin  
15
  39.1 or 140.9 
The value of  cannot be 149.0° as the total of the angles would be greater than 180°.
So   39.1 .
Angle A  180  52  39.08  88.92
x
15

Using the sine rule:
sin 88.92 sin 52
15 sin 88.92
x
 19.0 cm
sin 52
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EdExcel C2 Geometry & trig Assessment solutions
6.
x
B

O
25°
40°
25° 3
5
40°
A
(i) Using the cosine rule: x 2  5 2  3 2  2  3  5 cos 65
x  4.62 km
[3]
4.6175  3  5
2  3  4.6175
  78.9
Bearing  180  25  78.9
(ii) Using the cosine rule: cos  
2
2
2
 283.9
[4]
7. cos  
Adjacent
2.5 1


Hypotenuse
5
2
x
2.5

  60
Angle of arc with belt on = 360  60  60  240

4
240  240 

180
3
4  10
Arc length  r   2.5 

3
 3 
tan  
5
x
2.5
x  2.5 tan 60
10
 2x
3
10

 5 tan 60
3
 19.1 m(3 s.f.)
Total length of belt 
[5]
8. (i) If the triangle has an angle of 60 at the centre of the circle then it must be an
equilateral triangle and so part of the perimeter is 3cm.
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EdExcel C2 Geometry & trig Assessment solutions
60  60 

180


3

Arc length  r   3    
3
Perimeter  (  3) cm
 6.14 cm (3 s.f.)
3
3
2
Area  Area of sector  Area of triangle
3 1

  3  3 sin 60
2 2
 0.815 cm 2(3 s.f.)
Area of sector  21 r 2  0.5  32 


[7]
(ii) Arc length = r
Sector area = 21 r 2
45  45 
Sector 1:

180


4
Arc length  10 
Sector area 
Sector 2:
1
2
Arc length  8 

4

5
2
 10 2 

4

4

25 
2
 2
Sector area  21  8 2 

4
 8
Shaded area: Perimeter  2  2  Arc length 1  Arc length 2
5
 4
 2
2
 18.1 cm (3 s.f.)
Area  Area of sector 1  Area of sector 2
25 

 8
2
 14.1 cm 2(3 s.f.)
[7]
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