1. Coordinate Geometry 2 The Circle
27th November 2017

2. The Line – Key Concepts Revision Circle Question Focused Approach
Welcome!
Coordinate Geometry –
The Line – Key Concepts Revision
Circle
Question Focused Approach
2 hours - 10mins
Any questions? – Just Ask!!
Welcome: Likely last until (depending on when kids arrive).
Introduce tutors
The tutorial is going to last for 2 hours
Break in the middle with biscuits and water
As per previous tutorials –we’ll get them to try questions before we go through the solutions
Tutors will be walking around to help out
Safety precautions to be covered? The fire exits are…….

3. Useful Resource for Project Maths
Normally €49.99
Discount code: SOA2018
€29.99 for the year
Access to 11 June 2018

4. Material from tonight
Slides, questions and solutions available from website
Or google ‘actuaries maths tutorials’ to find it

5. Exam Technique:
Read the question carefully.
Key formula tables – page 18 & 19
Draw a diagram every time.
Label all diagrams.

6. Question 1 The coordinates of 3 points A, B and C are:
Find the equation of AB.
The line AB intersects the y-axis at D.
Find the coordinates of D.
Find the perpendicular distance from C to AB.
Hence, find the area of the triangle ADC.
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑥1,𝑦1 𝑓𝑟𝑜𝑚 𝑙𝑖𝑛𝑒 𝑎𝑥+𝑏𝑦+𝑐=0 𝑖𝑠
|𝑎𝑥1+𝑏𝑦1+𝑐| ( 𝑎 2 + 𝑏 2 )

7. Question 1 D
A (2,2)
C (-2,-3)
B (6,-6)

8. Question 1 Solution Find the slope using slope formula
𝑚= (𝑦2−𝑦1) (𝑥2−𝑥1) = (−6−2) (6−2) = −8 4 =−2
Slope = -2, A = (2,2)
Equation of AB: y - 2 = -2 * (x - 2)
y - 2 = -2x +4
2x + y = 6

9. Question 1
Solution
Line intersects the y-axis means that x=0 at this point
Substitute x=0 into Equation of AB: 2x + y = 6
y = 6
D = (0,6)
Solution
Formula in Tables Page 19
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒= |𝑎𝑥1+𝑏𝑦1+𝑐| ( 𝑎 2 + 𝑏 2 ) = |2 −2 +1 −3 −6| ( ) = |−13| =

10. Question 1
Solution
Area of Triangle is ½ x base x perpendicular height
Perpendicular height is from last question
Distance between A(2,2) and D(0,6) is the base
(0−2 ) 2 +( 6−2) 2
4+16 = 20
Area = ½ * 20 * = 13 units squared
Note: To use formula for area of triangle on page 18 it is necessary to have one apex at the origin.

11. Section 2: Circles Key formulae – page 19
Finding the equation of a circle:
(𝑥−ℎ) 2 + (𝑦−𝑘) 2 = 𝑟 Formula for circle with centre (h, k)
𝑥²+𝑦² =𝑟² Formula for circle with centre (0, 0)
Finding the centre and radius of a circle:
Express the circle in this form: 𝑥 2 + 𝑦 2 +2𝑔𝑥+2𝑓𝑦+𝑐=0
Centre of the circle is (-g, -f)
Radius of the circle is √(g2 + f2 – c)

12. Question 2 A circle has centre (2,3) and contains the point (8,9)
Sketch the circle
Find the radius length of the circle
Write down the equation of the circle

13. a) Solution y
(8,9)
(2,3)

14. Question 2 b) Solution Distance between (8,9) and (2,3) is radius
(2−8 ) 2 +( 3−9) 2
(−6) 2 + (−6) 2 = = 72
Solution
Equation of circle with centre (h,k) and radius r is:
(𝑥−ℎ) 2 + (𝑦−𝑘) 2 = 𝑟 2
(𝑥−2) 2 + (𝑦−3) 2 = =72

15. Question 3
The line segment joining A(-5,3) and B(5,-3) is the diameter of a circle
Sketch the circle
Find the centre of the circle
Find the radius length of the circle ( (𝑥2−𝑥1 ) 2 +( 𝑦2−𝑦1) 2 )
Write down the equation of the circle
Using the formula Area = π𝑟² find the area of the circle. Give your answer correct to two decimal places.
Find the area of the square in which the circle can be inscribed

16. a) Solution y
A(-5,3)
B(5,-3)

17. Question 3
b) Solution
Midpoint of A(-5,3) and B(5,-3) is the centre of the circle
Midpoint is (0,0)
Solution
Distance between (−5,3) and (0,0) is radius
(−5−0 ) 2 +( 3−0) 2 34

18. Question 3 Solution 𝑥²+𝑦² =𝑟² 𝑥²+𝑦² =( 34 )² 𝑥²+𝑦² =34 e) Solution
𝑥²+𝑦² =( 34 )²
𝑥²+𝑦² =34
e) Solution
Area = π𝑟²= π
34 π
units squared

19. Question 3(f) y
A(-5,3)
B(5,-3)

20. Question 3
f) Solution
Side of square is equal to the diameter, area of square is diameter ²
Diameter = 2r = 2 34
Area of square = ( ) ²
136 units squared

21. Sample paper 2012 –Q2 (25 marks)
Question 4
Sample paper 2012 –Q2 (25 marks)
The equations of two circles are:
c₁ : 𝑥²+𝑦²−6𝑥−10𝑦+29=0
c₂ : 𝑥²+𝑦²−2𝑥−2𝑦−43=0
Write down the centre and radius-length of each circle.
Prove that the circles are touching.
Verify that (4, 7) is the point that they have in common.
Find the equation of the common tangent.

22. Sample paper 2012 –Q2 (25 marks)
Solution circle c1
𝑥 2 + 𝑦 2 +2𝑔𝑥+2𝑓𝑦+𝑐=0
𝑥 2 + 𝑦 2 −6𝑥−10𝑦+29=0
2g = g = f = f = 5 (Formulae p. 19)
centre (3, 5)
Radius formula (p. 19) √(g2 + f2 – c), c = 29
−29 = −29 = 5

23. Sample paper 2012 –Q2 (25 marks)
Solution circle c2
𝑥 2 + 𝑦 2 −2𝑥−2𝑦−43=0
2g = g = f = f = 1
centre (1,1)
Radius formula
= = 9∗5 = 3 5

24. Sample paper 2012 –Q2 (25 marks)
Solution
Distance between centres:
( 3−1) 2 +( 5−1) 2 20
2 5
The distance between the centres is the difference of the radii => circles touch (internally).

25. Sample Paper 2012 – Q2
x2+y2-6x-10y+29=0
x2+y2-2x-2y-43=0

26. Sample paper 2012 –Q2 (25 marks)
Solution
Substitute (4,7) into c₁ and c₂
−6 4 − =0
(4, 7) is point on c₁
−2 4 −2 7 −43=0
(4, 7) is point on c₂
Circles touch internally at (4,7)

27. Sample Paper 2012 – Q2 (d)
(4,7)

28. Sample paper 2012 –Q2 (25 marks)
Solution
Slope from (3, 5) to (4, 7) is: 7−5 4−3 = 2
Slope of tangent = − 1 2 (as perpendicular)
Equation of tangent with slope − 1 2 and point(4,7)
𝑦 −7=− 𝑥−4
2𝑦 −14=−𝑥+4
𝑥 + 2𝑦 −18=0

29. Question 5
A circle passes through the point (3,3) and the point (4,1).
If the centre of the circle is on the line x + 3y = 12, find its equation.
NB: Draw a rough sketch of the circle and the line above

30. Question 5
The perpendicular bisector of any chord is a line containing the centre of the circle
(0,4)
Get students to sketch down the graph before attempting the Q
(3,3)
x + 3y = 12
(4,1)
(12,0) 30

31. Question 5
The perpendicular bisector of any chord is a line containing the centre of the circle
End points of chord: (3,3) and (4,1)
Midpoint of Chord: (3.5, 2)
slope of chord: (1-3)/(4-3) = -2
slope of perpendicular of chord: 1 2

32. Question 5 Eqn of perpendicular of chord: slope = 1 2 , point (3.5,2)
y−2= 1 2 (𝑥−3.5)
2y−4=𝑥−3.5
x−2y=−0.5
x+3y=12 point of intersection is centre of circle
−5y=−12.5 subtracting
y=2.5; 𝑥=4.5
Centre is (4.5,2.5)

33. Question 5 Now, need to find the radius
Distance between (4.5,2.5) and (3,3)
r= (3−4.5 ) 2 +( 3−2.5) 2
r= 2.5
𝑥 − 𝑦 − =2.5
𝑥 2 + 𝑦 2 −9𝑥−5𝑦+24 =0

34. Question 6 The line 3x -4y+14=0 is tangent to a circle at the point
(-2, 2). The circle also contains the point (5,1)
Draw a rough sketch of the circle.
Find the equation of the circle.

35. Question 6
3x -4y+14=0
(-2,2)
(5,1)

36. Question 6
Solution
Need to find the centre of the circle to find the equation of the circle
From Hint 4
The radius is perpendicular to the tangent at the point of intersection
The perpendicular bisector of any chord is a line containing the centre of the circle
The point of intersection of these two lines is the centre of the circle

37. Question 6 Tangent: 3x -4y+14=0 Slope of tangent = ¾
slope of perpendicular = -(4/3)
Equation of perpendicular: slope -4/3, point (-2,2)
y−2=−( 4 3 )(x−(−2))
3y−6=−4x−8
4x+3y=−2

38. Question 6 To find perpendicular bisector of chord:
Midpoint (-2,2) and (5,1): (1.5, 1.5)
Slope of (-2,2) and (5,1): (1-2)/(5-(-2))
Slope: -(1/7)
Slope of perpendicular = 7
Equation of perpendicular: slope 7, point (1.5,1.5)
y−1.5=7(x−1.5)
7x−y=9

39. Question 6 Simultaneous equations 7x−y=9 4x+3y=−2 21x−3y=27
Adding: 25x=25;
x=1 and y=−2
centre (1, −2)

40. Question 6
So, centre (1,-2)
Radius is distance between (1,-2) and (5,1)
r= (5−1 ) 2 +( 1−(−2)) 2
r= 16+9
r = 5
Equation of circle: (𝑥−1 ) 2 +( 𝑦−(−2)) 2 = 5 2
𝑥 2 + 𝑦 2 −2𝑥+4𝑦−20=0

41. Monday 4th December Statistics Same location 6 - 8 pm Next tutorial