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Coordinate Geometry of the line questions The circle C has the equation x 2 + y 2 = 49 (i) Write down the centre and radius of C? Centre = (0,0) Radius = √49 = 7 (ii) Verify that the point (5, -5) lies outside C (5)2 + (-5)2 = 25 + 25 = 50 > 49 so it is outside (b) The Line y = 10 – 2x intersects with the circle x 2 + y 2 = 40 Find the points of intersection? • x 2 + (10 – 2x) 2 = 40 • x 2 + 100 – 40x + 4x 2 = 40 • 5x 2 – 40x + 60 = 0 • x – 8x + 12 = 0 •(x – 6)(x – 2) = 0 •Solutions: x = 6 or 2 •Corresponding y values? • y = 10 – 2(6) = -2 • y = 10 – 2(2) = 6 •Answers: (6 , -2) and (2 , 6) 2002 LC Paper 2 Q. 3 (0, 10) 10 (-10, 0) 10 (10, 0) 10 10 x2 + y2 = 100 Centre (0, 0) |Radius| = 10 Points on circle: (10, 0), (0, 10) (-10, 0) Equation of Circle C: (x - 2)2 + (y + 1)2 = 8 Centre (2, -1) |Radius| = 8 C cuts y-axis when x = 0 (y + 3)(y – 1) = 0 (0 - 2)2 + (y + 1)2 = 8 y + 3 = 0 or y – 1 = 0 4 + y2 + 2y + 1 = 8 y2 y2 + 2y + 5 - 8 = 0 + 2y - 3 = 0 y = - 3 or y = 1 C cuts y-axis at a(0, -3) and b(0, 1) (x – 2)2 + (y + 1)2 = 8 T Centre (2, -1) |Radius| = 8 RT y2 – y1 Slope of R = x2 – x1 x1 = 2 x2 = 4 y1 = -1 y2 = 1 (1) – (-1) (2, -1) R (4, 1) = (4) – (2) 2 = 2 =1 Slope of T = - 1 Point on T (4, 1) Slope of T = -1 Equation of T: y – y1 = m(x – x1) x1 = 4 m = -1 y1 = 1 y – (1) = -1(x – (4)) y–1=-x+4 x + y – 5 = 0 Equation of T 8 b(3, 7) 7 6 5 4 3 2 a(-5, 1) 1 0 -6 -4 -2 0 2 4 6 8 -1 -2 From the diagram the right-angle appears to be at b We have to show that ab bc 10 c(9, -1) y2 – y1 Slope of ab = = = x2 – x1 (7) – (1) (3) – (-5) 6 8 3 = 4 x1 = - 5 y1 = 1 x2 = 3 y2 = 7 Slope of bc = = = y2 – y1 x2 – x1 (-1) – (7) (9) – (3) -8 6 -4 = 3 x1 = 3 y1 = 7 x2 = 9 y2 = -1 Slope of ab = 3 4 Product of slopes = Slope of bc = 3 x 4 4 3 ab bc abc is right-angled at b = -1 -4 3 b(3, 7) Centre of Circle, d, is the mid-point of [ac] |Radius| = |ad| a(-5, 1) d c(9, -1) abc = 900 is an angle in a semi-circle of which [ac] is the diameter a(-5, 1) d(2, 0) c(9, -1) x1 x2 y1 y2 Mid-point of [ac] = 2 , 2 x1 = - 5 x2 = 9 y1 = 1 y2 = -1 (5) (9) (1) (1) , 2 2 5 9 1 1 , 2 2 d 2, 0 = Centre of Circle |Radius| = Distance from d(2, 0) to a(-5, 1) x2 x1 2 y2 y1 2 | ac| x1 = -5 x2 = 2 y1 = 1 y2 = 0 2 (-5) 2 0 12 7 2 12 49 1 50 = |Radius| Centre = (2, 0) |Radius| 50 Equation of Circle: (x - 2)2 + (y)2 = 50 2001 Paper 2 Question 3 S: (x – 3)2 + (y – 4)2 = 25 Centre: (3, 4) |Radius| = 5 (k, 0) is on the circle S k(k – 6) = 0 (k – 3)2 + (0 – 4)2 = 25 k = 0 or k - 6 = 0 k2 – 6k + 9 + 16 = 25 k = 0 or k = 6 k2 – 6k + 25 – 25 = 0 k2 – 6k = 0 The line is a tangent if there is only one point of intersection x – 3y = 10 x2 + y2 = 10 Substitute 3y + 10 for x x = 3y + 10 (3y + 10)2 + y2 = 10 9y2 + 60y + 100 + y2 = 10 10y2 + 60y + 90 = 0 Divide by 10 y2 + 6y + 9 = 0 The line is a tangent if there is only one point of intersection x – 3y = 10 y2 + 6y + 9 = 0 x = 3y + 10 (y + 3)(y + 3) = 0 When y = -3 x = 3(-3) + 10 x = - 9 + 10 = 1 y + 3 = 0 or y + 3 = 0 y=-3 Only one point of intersection (1, -3) The line is a tangent o(0, 0) a(1, -5) | oa | x2 x1 2 y2 y1 2 1 02 - 5 02 12 - 52 Centre = (0, 0) |Radius| 26 Equation of C: x2 + y2 = 26 1 25 26 x1 = 0 x2 = 1 y1 = 0 y2 = -5 = |radius| (p, p) First Component = Second Component is on the line y = x 6 5 PZ (4,4) 4 (3,3) 3 (2,2) 2 1 -6 -5 -4 -3 0 (0,0) -1 (-1,-1) 0 1 -1 -2 (-2,-2) -2 (-3,-3) (-4,-4) -3 -4 -5 -6 (1,1) 2 3 4 5 6 Values of p are -3, -2, -1, 0, 1, 2, 3 2000 Paper 2 Question 3 x2 + y2 = 16 is the circle C Centre (0, 0) and |Radius| = 4 |Radius | = 4 o(0, 0) a(3, 1) x2 x1 2 y2 y1 2 | oa | 3 02 1 02 32 12 9 1 10 = 3.16 < |Radius| a(3,1) is inside the circle x1 = 0 x 2 = 3 y1 = 0 y2 = 1 x2 + y2 = 29 Centre (0, 0) |Radius| = 29 T (0, 0) R (2, 5) RT y2 – y1 Slope of R = x2 – x1 = (5) – (0) (2) – (0) 5 = 2 5 = 2 2 Slope of T= - 5 x1 = 0 x2 = 2 y1 = 0 y2 = 5 Point on T (2, 5) 2 Slope of T= - 5 Equation of T: y – y1 = m(x – x1) x1 = 2 y1 = 5 m = -2 5 y – (5) = -2 (x – (2)) 5 y - 5 = -2 (x - 2) 5 Multiply by LCD = 5 5y - 25 = (5)(-2) (x - 2) = - 2x + 4 5 2x + 5y - 29 = 0 Equation of T a(-2, -3) b(-4, 3) Centre of Circle is the mid-point of the diameter ab a(-2, -3) c(-3, 0) b(- 4, 3) x1 x2 y1 y2 Mid-point of [ab] = 2 , 2 x1 = - 2 x2 = - 4 y1 = - 3 y2 = 3 (2) (4) (3) (3) , 2 2 2 4 3 3 6 , 0 c - 3, 0 , 2 2 2 2 = Centre of Circle |Radius| = Distance from c(-3, 0) to a(-2, -3) | ac| x2 x1 2 y2 y1 2 - 3 - 22 0 - 32 - 3 22 0 32 - 12 32 19 10 = |Radius| x1 = -2 x2 = -3 y1 = -3 y2 = 0 Centre = (-3, 0)|Radius| 10 Equation of Circle: (x + 3)2 + (y)2 = 10 Equation of Circle: (x + 3)2 + y2 = 10 Circle cuts y-axis when x = 0 (0 + 3)2 + y2 = 10 9 + y2 = 10 y2 = 10 – 9 = 1 y2 –1=0 (y + 1)(y – 1) = 0 y + 1 = 0 or y – 1 = 0 y = - 1 or y = 1 Circle cuts y-axis at a(0, -1) and b(0, 1) These points on the y-axis are 2 units apart |ab| = 2 From an Accurate diagram 3 2 6 c -6 b(0, 1) 1 -5 -4 20 -3 -2 -1 0 -1 d Area of•abcd = (6)(2) = 12 -2 -3 1 a(0, -1) 2 3 1999 Paper 2 Question 3 o(0, 0) a(8, 6) x2 x1 2 y2 y1 2 | oa | 8 0 2 6 0 2 82 62 64 36 100 10 = |Radius| x1 = 0 x2 = 8 y1 = 0 y2 = 6 Centre = (0, 0) |Radius| = 10 Equation of C: x2 + y2 = 100 a(-1, -1) b(3, -3) Centre of S is the mid-point of the diameter ab a(-1, -1) c(1, -2) b(3, -3) x1 = - 1 x2 = 3 y1 = - 1 y2 = - 3 Mid-point of [ab] x1 x2 y1 y2 = 2 , 2 1 3 1 (3) , 2 2 1 3 1 3 c 1,2 , 2 2 = Centre of S |Radius| = Distance from c(1, -2) to a(-1, -1) | ac| x2 x1 2 y2 y1 2 - 1 12 - 2 (1)2 - 22 - 2 12 Centre = (1, -2)|Radius| 5 Equation of S: (x – 1)2 + (y + 2)2 = 5 4 1 5 x1 = 1 x2 = -1 y1 = -2 y2 = -1 = |Radius| x2 + y2 = 13 Centre (0, 0) |Radius| = 13 T RT y2 – y1 Slope of R = x2 – x1 (-3) – (0) (0, 0) R (-2, -3) = (-2) – (0) -3 = -2 3 = 2 2 Slope of T= - 3 x1 = 0 x2 = -2 y1 = 0 y2 = -3 2 Point on T (-2, -3) Slope of T= - 3 Equation of T: y – y1 = m(x – x1) x1 = -2 y1 = -3 m = -2 3 y – (-3) = -2 (x – (-2)) 3 y + 3 = -2 (x +2) 3 Multiply by LCD = 3 3y + 9 = (3)(-2) (x +2) = - 2x - 4 3 2x + 3y + 13 = 0 Equation of T T (0, 0) 2 Slope= - 3 R The point needed is (-2, -3) the other end of the diameter as shown 2 Slope= - 3 The other end of the diameter is (2, 3) (0, 0) T (-2, -3) The translation that maps (-2,-3) (0, 0) will map (0, 0) ( 2,3 ) Point (2, 3) 2 Slope of T= - 3 Equation of || tangent: y – y1 = m(x – x1) x1 = 2 y1 = 3 m = -2 3 y – (3) = -2 (x – (2)) 3 y - 3 = -2 (x - 2) 3 Multiply by LCD = 3 3y - 9 = (3)(-2) (x - 2) = - 2x + 4 3 2x + 3y - 13 = 0 Equation of other tangent