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C2: Coordinate geometry Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 14th September 2013 Midpoint of two points and distance between them Midpoint Distance (1,1) (2,2) (1.5,? 1.5) √2 ? (1,2) (4,6) ? 4) (2.5, 5? (1,-4) (-3,5) (-1,?0.5) √97 ? (-1,-2) (-2,10) ? 4) (-1.5, ? √145 (a,a-1) (b,a+1) ? ([a+b]/2, a) √[(b-a)? 2 +4] Centres of circles The line PQ is a diameter of the circle centred at (2, -2). Given P is (8, -5), find the coordinate of Q. Q = (-4, 1) ? The points M(3, p) and N(q, 4) lie on the circle centre (5, 6). The line MN is a diameter of the circle. Find the value of p and q. p = 8, q? = 7 Radii of circles The line PQ is a diameter of a circle, where P and Q are (-1, 3) and (6, -3) respectively. Find the radius of the circle. r = 0.5√85 ? The points (-3, 19), (-15, 1) and (9, 1) are points on the circumference of a circle. Show that (-3, 6) is the centre of the circle. We just need to show that each point is equidistant from the centre. ? centre turns out to be 13. The distance for all three points to the Lines through circles The line AB is a diameter of the circle centre C, where A and B are (-1, 4) and (5,2) respectively. The line l passes through C and is perpendicular to AB. Find the equation of l. y = 3x -? 3 The line PQ is a chord of the circle centre (-3, 5), where P and Q are (5, 4) and (1, 12) respectively. The line l is perpendicular to PQ and bisects it. Prove that l passes through the centre of the circle. Equation of perpendicular line: = ? When = 3, = −5 1 2 + 13 2 Lines through circles The perpendicular from the centre of a circle to a chord bisects the chord. We say that the radius is the perpendicular bisector of the chord. Lines through circles y D x The lines and are chords of a circle. The line = 3 − 11 is the perpendicular bisector of . The line = − − 1 is the perpendicular bisector of . Find the coordinates of the centre of the circle. C , ?− A B Proving a triangle has a right angle Suppose we know the three sides of a triangle. How do we prove it has a right angle? 5 3 It satisfies Pythagoras’ Theorem. ? 4 C How therefore could we prove that is the diameter of the circle? A B If AC2 + BC2 = AB2, then angle ACB is a right angle. Thus (from GCSE circle theorems!) AB must?be the diameter of the circle. Proving a triangle has a right angle If PR is the diameter, how would you prove that = 13? PQ2 = 208 QR2 = (a-9)2 + 36 PR2 = (a+3)2 + 4 ? Setting PQ2 + QR2 = PR2 and solving, we find that a = 13. Questions 1 Ex4A Q10) The points V(-4, 2a) and W(3b, -4) lie on the circle centre (b, 2a). The line VW is a diameter of the circle. Find the value of a and b. 3 a = -2, b ? =4 2 Ex4B Q1) The line FG is a diameter of the circle centre C, where F and G are (-2, 5) and (2, 9) respectively. The line l passes through C and is perpendicular to FG. Find the equation of l. y= -x + 7 ? 4 Ex4B Q10) Find the centre of a circle with points on its circumference of (-3,19), (9,11) and (-15, 1). (-3, 6) ? Ex4B Q9) The points P(3, 16), Q(11, 12) and R(-7, 6) lie on the circumference of a circle. a) Find the equation of the perpendicular bisector of: i. PQ ii. PR b) Hence, find the coordinates of the centre of the circle. a) i) y = 2x ii) y =?-x + 9 5 b) (3, 6) Ex4C Q9) The points A(2, 6), B(5,7) and C(8,2) lie on a circle. a) Show that triangle ABC has a right angle. b) Find the area of the triangle. c) Find the centre of the circle. ? ? ? a) AB = √10, AC = 10. BC = r90. AB2 + BC2 = AC2. b) 15 c) (5, 2) Equation of a circle y What is the equation of a circle with radius r, centred at the origin? (x,y) r r x (Hint: draw a right-angled triangle inside your circle, with one vertex at the origin and another at the circumference) y r x x2 + ?y2 = r2 Equation of a circle y Now suppose we shift the circle so it’s now centred at (a,b). What’s the equation now? r (Hint: What would the sides of this rightangled triangle be now?) (a,b) x ? 2 = r2 (x-a)2 + (y-b) Equation of a circle Centre Radius Equation (0,0) 5 x2 + y?2 = 25 (1,2) 6 (x-1)2 + (y-2) ? 2 = 36 (-3,5) ? ?1 ?7 ?4 √3 ? 2√2 ? (x+3)2 + (y-5)2 = 1 (-5,2) ? (-6,-7) ? (1,-1) ? (-2,3) ? (x+5)2 + (y-2)2 = 49 (x+6)2 + (y+7)2 = 16 (x-1)2 + (y+1)2 = 3 (x+2)2 + (y-3)2 = 8 Equation of a circle ? Show that the circle (x-3)2 + (y+4)2 = 20 passes through (5, -8). Just substitute x=5 and y=-8 and show ? the equation holds! ? Where does the circle (x – 1)2 + (y – 3)2 = 45 meets the x-axis? On x-axis, y = 0. So (x – 1)2 + 9 = 45.?Solving gives (7,0) and (-5,0) ? Given that AB is the diameter of a circle where A and B are (4,7) and (-8,3) respectively, find the equation of the circle. (x+2)2 + (y-5)2 = 400 ? ? The line 4x – 3y – 40 = 0 is a tangent to the circle (x – 2)2 + (y – 6)2 = 100 at the point P(10,0). Show that the radius at P is perpendicular to the line (as we would expect!) Equation of line is y = 4/3 x – 40/3, so gradient is 4/3 Centre of circle is (2,6) So gradient of radius is (6-0)/(2-10) = -3/4. ? Since 4/3 x -3/4 = -1, radius is perpendicular to tangent. P Intersections! How could you tell if a line and a circle intersect: 0 times once twice (x+2)2 + y2 = 33 Equate the expressions then look at the discriminant: b2 – 4ac ? <0 b2 – 4ac ? >0 b2 – 4ac ? =0 This allows us to prove that a line is a tangent to the circle. Completing the square [Edexcel] The circle C, with centre at the point A, has equation x2 + y2 – 10x + 9 = 0. Find (a) the coordinates of A, (2) (b) the radius of C, (2) x2 – 10x + y2 + 9 = 0 (x – 5)2 – 25 + y2 + 9 = 0 (x – 5)2 + y2 = 16 ? So centre is (5,0), radius is 4. Equation of a circle Equation Centre x2 + y2 + 4y – 5 = 0 (0,?-2) x2 + y2 – 6x + 4y – 3 = 0 (3,?-2) x2 + y2 + 12x + 2y + 12 = 0 (-6,?-1) x2 + y2 – 4x – 6y = 3 (2,?3) x2 + y 2 + x + y = 1 (-0.5,?-0.5) Radius ?3 ?4 ?5 ?4 ? 3/2 (Note: this appears in exams, but not in your textbook!)