Coordinate Geometry 2 The Circle 27th November 2017  

The Line – Key Concepts Revision Circle Question Focused Approach Welcome! Coordinate Geometry – The Line – Key Concepts Revision Circle Question Focused Approach 2 hours - 10mins break @7pm Any questions? – Just Ask!! Welcome: Likely last until 18.05- 18.10 (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…….

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

Material from tonight Slides, questions and solutions available from website https://web.actuaries.ie/students/tutorials Or google ‘actuaries maths tutorials’ to find it

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

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 )

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

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

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| ( 2 2 + 1 2 ) = |−13| 5 = 13 5

Question 1 Solution Area of Triangle is ½ x base x perpendicular height Perpendicular height is 13 5 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 5 = 13 units squared Note: To use formula for area of triangle on page 18 it is necessary to have one apex at the origin.

Section 2: Circles Key formulae – page 19 Finding the equation of a circle: (𝑥−ℎ) 2 + (𝑦−𝑘) 2 = 𝑟 2 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)

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

a) Solution y (8,9) (2,3)  

Question 2 b) Solution Distance between (8,9) and (2,3) is radius (2−8 ) 2 +( 3−9) 2 (−6) 2 + (−6) 2 = 36+36 = 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 2 =72

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

a) Solution y A(-5,3)   B(5,-3)

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 5 2 + 3 2 34

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 2 34 π 106.81 units squared

Question 3(f) y A(-5,3)   B(5,-3)

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 34 ) ² 136 units squared

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.

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 = -6 -g = 3 2f = -10 -f = 5 (Formulae p. 19) centre (3, 5) Radius formula (p. 19) √(g2 + f2 – c), c = 29 3 2 + 5 2 −29 = 9+25 −29 = 5

Sample paper 2012 –Q2 (25 marks) Solution circle c2 𝑥 2 + 𝑦 2 −2𝑥−2𝑦−43=0 2g = -2 -g = 1 2f = -2 -f = 1 centre (1,1) Radius formula 1 2 + 1 2 +43 = 45 = 9∗5 = 3 5

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).

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

Sample paper 2012 –Q2 (25 marks) Solution Substitute (4,7) into c₁ and c₂ 4 2 + 7 2 −6 4 −10 7 +29=0 (4, 7) is point on c₁ 4 2 + 7 2 −2 4 −2 7 −43=0 (4, 7) is point on c₂ Circles touch internally at (4,7)

Sample Paper 2012 – Q2 (d)   (4,7)  

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=− 1 2 𝑥−4 2𝑦 −14=−𝑥+4 𝑥 + 2𝑦 −18=0

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

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

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

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)

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 −4.5 2 + 𝑦 −2.5 2 =2.5 𝑥 2 + 𝑦 2 −9𝑥−5𝑦+24 =0

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.

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

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

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

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

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

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

Monday 4th December Statistics Same location 6 - 8 pm Next tutorial https://web.actuaries.ie/students/tutorials