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Higher Mathematics

Higher Mathematics. Circle. Unit 2 Outcome 4. Centre O (0,0) and radius r. x 2 + y 2 = r 2. DiscoveringTime. Friday, 07 November 2014. y. P( x , y ). r. O. Q. x. The equation of a circle centred on the origin.

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Higher Mathematics

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  1. Higher Mathematics Circle Unit 2 Outcome 4 Centre O(0,0)and radiusr x2 + y2 = r2 DiscoveringTime Friday, 07 November 2014

  2. y P(x, y) r O Q x The equation of a circle centred on the origin Suppose we have a circle with its centre on the origin O and radius r. If P(x, y) is any point on the circle we can write OP = r. Let’s add another point Q on the x-axis so OQ = x and PQ = y. y Using Pythagoras’ theorem: x OQ2 + PQ2 = OP2 So x2 + y2 = r2 The equation of a circle of radius r centred on the origin is x2 + y2 = r2

  3. Higher Mathematics Circle Unit 2 Outcome 4 Centre O(0,0)and radiusr x2 + y2 = r2 State the radius of each circle x2 + y2 = 49 7 x2 + y2 = 64 8 x2 + y2 = 81 9 x2 + y2 = 16 4 x2 + y2 = 36 6 3 x2 + y2 = 9 1 x2 + y2 = 4 2 x2 + y2 = 1 x2 + y2 = 100 10 x2 + y2 = 144 12 discoveringtime.com Friday, 07 November 2014

  4. Higher Mathematics Circle Unit 2 Outcome 4 Centre O(0,0)and radiusr x2 + y2 = r2 Write the equation of the circle with the centre (0,0) and passing through a) (6,8) c) (-2,1) 62 + 82 = r2 -22 + 12 = r2 36 + 64 = r2 4 + 1 = r2 100 = r2 5 = r2 x2 + y2 = 100 x2 + y2 = 5 b) (3,5) d) (0, -3) 32 + 52 = r2 02 + -32 = r2 9 + 25 = r2 0 + 9 = r2 34 = r2 9 = r2 x2 + y2 = 34 x2 + y2 = 9 DiscoveringTime Friday, 07 November 2014

  5. Higher Mathematics Circle Unit 2 Outcome 4 Centre O(0,0)and radiusr x2 + y2 = r2 Write the equation of the circle with a centre origin and radius c) 9 a) 4 r= 4 r= 9 r2 = 16 r2 = 81 x2 + y2 = 16 x2 + y2 = 81 b) 7 d) 12 r= 12 r= 7 r2 = 144 r2 = 49 x2 + y2 = 144 x2 + y2 = 49 DiscoveringTime Friday, 07 November 2014

  6. Higher Mathematics Circle Unit 2 Outcome 4 Find the centre and radius of the circles below x2 + y2 = 7 x2 + y2 = 11 centre (0,0) centre (0,0) radius = 7 radius = 11 x2 + y2 = 64 x2 + y2 = 25 centre (0,0) centre (0,0) radius = 8 radius = 5 DiscoveringTime Friday, 07 November 2014

  7. Higher Mathematics Unit 2 Outcome 4 To build skills Complete Page 168 Exercise 1 DiscoveringTime Friday, 07 November 2014

  8. Higher Mathematics Unit 2 Outcome 4 Question 5 Question 1 Question 2 f) Remember to get the equation in correct format. Divide by 3 Draw a diagram Part ii) radius is the hypotenuse, use Pythagoras e) (sqrt 3 )2 = 3 Question 6 Question 7 2 values when you do a square root !!! Using skills. Method same as before Question 10 Question 11 Question 8 Problem solving Draw diag Use pythagoras Take care with layout. Ratio problem first Given Remember to state final answer 1 sec radius 2cm so 2sec radius 4cm etc Plenary question …… DiscoveringTime Friday, 07 November 2014

  9. Higher Mathematics Circle Unit 2 Outcome 4 y x Exam standard question A square emblem has a circle at its centre with equation x2 + y2 = 4 The pattern repeats to the edge of the emblem with circles on the same centre with a radius of 1cm more each time. The emblem side is 10 cm what is the equation of the largest circle on the emblem? Answer Radius of first circle is 2cm diam 4cm diam of second circle is 6cm diam of third circle is 8cm diam of fourth circle is 10cm with a radius 5 cm x2 + y2 = 25 DiscoveringTime Friday, 07 November 2014

  10. Higher Mathematics Circle Unit 2 Outcome 4 DiscoveringTime Friday, 07 November 2014

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