# C2: Coordinate geometry Dr J Frost Last modified: 14 th September 2013.

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C2: Coordinate geometry Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 14 th September 2013

Midpoint of two points and distance between them (1,1) (2,2)(1.5, 1.5) √2√2 MidpointDistance (1,2) (4,6) 5 (2.5, 4) (a,a-1) (b,a+1)([a+b]/2, a) √ [(b-a) 2 +4] (1,-4) (-3,5)(-1, 0.5) √ 97 (-1,-2) (-2,10)(-1.5, 4) √ 145 ?? ?? ?? ?? ??

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. The distance for all three points to the centre turns out to be 13. ?

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. ?

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 D C B A x y ?

Proving a triangle has a right angle 5 3 4 Suppose we know the three sides of a triangle. How do we prove it has a right angle? It satisfies Pythagoras’ Theorem. A B C ? ?

Proving a triangle has a right angle PQ 2 = 208 QR 2 = (a-9) 2 + 36 PR 2 = (a+3) 2 + 4 Setting PQ 2 + QR 2 = PR 2 and solving, we find that a = 13. ?

Questions 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. a = -2, b = 4 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 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 b) (3, 6) Ex4B Q10) Find the centre of a circle with points on its circumference of (-3,19), (9,11) and (-15, 1). (-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. AB 2 + BC 2 = AC 2. b)15 c)(5, 2) 1 2 3 4 5 ? ? ? ? ? ? ?

Equation of a circle x y r What is the equation of a circle with radius r, centred at the origin? (Hint: draw a right-angled triangle inside your circle, with one vertex at the origin and another at the circumference) x 2 + y 2 = r 2 x y r ? r (x,y)

Equation of a circle x y Now suppose we shift the circle so it’s now centred at (a,b). What’s the equation now? (Hint: What would the sides of this right- angled triangle be now?) (x-a) 2 + (y-b) 2 = r 2 ? (a,b) r

Equation of a circle CentreRadiusEquation (0,0)5x 2 + y 2 = 25 (1,2)6(x-1) 2 + (y-2) 2 = 36 (-3,5)1(x+3) 2 + (y-5) 2 = 1 (-5,2)7(x+5) 2 + (y-2) 2 = 49 (-6,-7)4(x+6) 2 + (y+7) 2 = 16 (1,-1) √3√3 (x-1) 2 + (y+1) 2 = 3 ? ? ?? ?? ?? ?? (-2,3) 2√ 2 (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! 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!) P 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. ? ? ? 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) ? ? ? ? ?

Intersections! How could you tell if a line and a circle intersect: 0 times twice once Equate the expressions then look at the discriminant: b 2 – 4ac < 0 b 2 – 4ac > 0b 2 – 4ac = 0 ? ?? (x+2) 2 + y 2 = 33 y = x - 7 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 x 2 + y 2 – 10x + 9 = 0. Find (a)the coordinates of A, (2) (b)the radius of C, (2) x 2 – 10x + y 2 + 9 = 0 (x – 5) 2 – 25 + y 2 + 9 = 0 (x – 5) 2 + y 2 = 16 So centre is (5,0), radius is 4. ?

Equation of a circle EquationCentreRadius x 2 + y 2 + 4y – 5 = 0(0, -2)3 ? x 2 + y 2 – 6x + 4y – 3 = 0(3, -2)4 x 2 + y 2 + 12x + 2y + 12 = 0(-6, -1)5 ? ? ? ? ? (Note: this appears in exams, but not in your textbook!) x 2 + y 2 – 4x – 6y = 3(2, 3)4 ?? x 2 + y 2 + x + y = 1(-0.5, -0.5) ??