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Higher Maths  2  4  Circles UNIT OUTCOME SLIDE
Distance Between Two Points NOTE SLIDE Higher Maths  2  4  Circles UNIT OUTCOME The Distance Formula d   =  (   y 2   –   y 1 ) Β² +  (   x 2   –   x 1 ) Β² √ B (   x 2   ,   y 2   ) A (   x 1   ,   y 1   ) y 2  –  y 1   x 2  –  x 1   Example Calculate the distance between (-2,9) and (4,-3). d   =  +  6 Β² √ 12 Β² = 180 √ = 5 √ 6 Where required, write answers as a  surd  in its simplest form. REMEMBER
Points on a Circle NOTE SLIDE Higher Maths  2  4  Circles UNIT OUTCOME Example Plot the following points and find a rule connecting  x   and  y . x y (   5   ,   0   ) (   4   ,   3   ) (   3   ,   4   ) (   0   ,   5   ) (-3   ,   4   ) (-4   ,   3   ) (-5   ,   0   ) (-4   ,-3   ) (-3   ,-4   ) (   0   ,-5   ) (   3   ,-4   ) (   4   ,-3   ) All points lie on a circle with radius 5 units and centre at the origin. x   Β²   +   y   Β²  =  25 x   Β²   +   y   Β²  =  r   Β² For any point on the circle, For any radius... NOTICE
The Equation of a Circle with centre at the Origin NOTE SLIDE Higher Maths  2  4  Circles UNIT OUTCOME x   Β²   +   y   Β²  =  r   Β² For any circle with radius  r  and centre the origin,  x y The β€˜Origin’ is the point  (0,0)   origin Example Show that the point  ( - 3   ,  ) lies on the circle with equation 7 x   Β²   +   y   Β²  =  16 x   Β²   +   y   Β² =  ( -3 ) Β²  +   (   ) Β² 7 =  9  +   7 =  16 Substitute point into equation: The point lies on the circle. LEARN THIS
The Equation of a Circle with centre  (   a   ,   b   ) NOTE SLIDE Higher Maths  2  4  Circles UNIT OUTCOME (   x   –   a   )   Β²   +  (   y   –   b   )   Β²   =  r   Β² For any circle with radius  r  and centre at the point  (   a   ,   b   )   ... Not all circles are centered at the origin. x y (   a   ,   b   ) r Example Write the equation of the circle with centre  (   3   , -5   ) and radius  2  3  . (   x   –   a   )   Β²   +  (   y   –   b   )   Β²   =  r   Β² (   x   –   3   )   Β²   +  (   y   –   ( -5 )   )   Β²   =  (  )   Β² 2  3 (   x   –   3   )   Β²   +  (   y   +   5   )   Β²   =  12 LEARN THIS
NOTE SLIDE Higher Maths  2  4  Circles UNIT OUTCOME The General Equation of a Circle (   x   +   g   ) 2   +  (   y   +   f   ) 2   =   r   2 (   x   2   +  2 g   x  +  g   2   )   +  (   y   2   +  2 fy  +  f   2   )   =   r   2 x   2   +  y   2   +  2 g   x   +  2 f   y   +   g   2  +  f   2  –  r   2  =   0 x   2   +  y   2  +  2 g   x  +  2 f   y   +  c   =   0 c   =   g   2   +   f   2   –   r   2 r   2   =   g   2   +   f   2   –   c   r   =   g   2   +   f   2   –   c   Try expanding the equation of a circle with centre  (   - g   ,   -   f  )   . General Equation of a Circle with center  (   - g   ,   -   f  ) and radius r   =   g   2   +   f   2   –   c   this is just a number... LEARN
NOTE SLIDE Higher Maths  2  4  Circles UNIT OUTCOME Circles and Straight Lines A line and a circle can have two, one or no points of intersection. r A line which intersects a circle at only one point is at  90Β° to the radius  and is is called a  tangent . two points of intersection one point of intersection no points of intersection REMEMBER
NOTE SLIDE Higher Maths  2  4  Circles UNIT OUTCOME Intersection of a Line and a Circle Example Find the intersection of the circle and the line 2   x   –  y  = 0 x   2   +  ( 2   x ) 2   =  45 x   2   + 4   x   2   =  45 5   x   2   =  45 x   2   =  9 x   =  3   or  -3 y  = 2   x   x   2   +  y 2   =  45 Substitute into  y  = 2   x  : How to find the points of intersection between a  line  and a circle: β€’  rearrange the equation of the line into the form  y  =   m   x   +   c  β€’ substitute  y  =   m   x   +   c   into the equation of the circle  β€’ solve the quadratic for  x  and substitute into  m   x   +   c   to find  y y   =  6   or  -6 Points of intersection are  ( 3,6 )  and  ( -3,-6 ) .
NOTE SLIDE Higher Maths  2  4  Circles UNIT OUTCOME Intersection of a Line and a Circle (continued) Example 2 Find where the line  2   x   –  y  + 8   =   0   intersects the circle  x   2   +  y   2  +  4   x  +  2   y  – 20   =   0 x y x   2   +  ( 2   x   +   8 ) 2  +  4   x  +  2   ( 2   x   +   8 )   – 20  =  0 x   2   +  4   x   2  +  32   x  +  64 + 4   x  + 4   x   + 16 – 20   =  0 5   x   2  +  40   x   + 60   =  0 5 (   x   2  +  8   x  +  12   )   =  0 5 (   x  +  2 )(   x  +  6 )   =  0 x   = -2   or   -6 Substituting into  y  = 2   x   +   8  points of intersection as ( -2,4 )  and  ( -6,-4 ) . Factorise  and solve
NOTE SLIDE Higher Maths  2  4  Circles UNIT OUTCOME The Discriminant and Tangents x   = - b b   2   –   ( 4   ac   ) Β± 2   a b   2   –   ( 4   ac   ) Discriminant The  discriminant  can be used to show that a line is a tangent: β€’  substitute  into the circle equation  β€’ rearrange to form a quadratic equation  β€’ evaluate the  discriminant y  =   m   x   +   c b   2   –   ( 4   ac   )   > 0 Two points of intersection b   2   –   ( 4   ac   )   = 0 The line is a  tangent b   2   –   ( 4   ac   )   < 0 No points of intersection  r REMEMBER
NOTE SLIDE Higher Maths  2  4  Circles UNIT OUTCOME Circles and Tangents Show that the line  3   x   +  y   =   -10   is a tangent to the circle  x   2   +  y   2  –  8   x  +  4   y  – 20   =   0 Example x   2   +  (- 3   x   –   10 ) 2  –  8   x  +  4   (- 3   x   –   10 )   – 20  =  0 x   2   +  9   x   2  +  60   x  +   100   –   8   x   –   12   x   – 40   – 20  =  0 10   x   2  +  40   x  +   40  =  0 b   2   –  ( 4   ac   ) =  40   2   –  (   4 Γ— 10 Γ— 40   ) =  0 =  1600   – 1600   The line is a tangent to the circle since b   2   –  ( 4   ac   )   =  0 x y
NOTE SLIDE Higher Maths  2  4  Circles UNIT OUTCOME Equation of Tangents To find the equation of a tangent to a circle: β€’ Find the center of the circle and the point where the tangent intersects β€’ Calculate the  gradient of the radius  using the gradient formula β€’ Write down the  gradient of the tangent β€’ Substitute the gradient of the tangent and the point of intersection into y   –   b   =   m   (   x   –   a   )   REMEMBER Straight Line Equation y   –   b   =   m   (   x   –   a   )   m   tangent   =   – 1 m   radius x 2   –   x 1   y 2   –   y 1   m   radius   = r

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Circles

  • 1. Higher Maths 2 4 Circles UNIT OUTCOME SLIDE
  • 2. Distance Between Two Points NOTE SLIDE Higher Maths 2 4 Circles UNIT OUTCOME The Distance Formula d = ( y 2 – y 1 ) Β² + ( x 2 – x 1 ) Β² √ B ( x 2 , y 2 ) A ( x 1 , y 1 ) y 2 – y 1 x 2 – x 1 Example Calculate the distance between (-2,9) and (4,-3). d = + 6 Β² √ 12 Β² = 180 √ = 5 √ 6 Where required, write answers as a surd in its simplest form. REMEMBER
  • 3. Points on a Circle NOTE SLIDE Higher Maths 2 4 Circles UNIT OUTCOME Example Plot the following points and find a rule connecting x and y . x y ( 5 , 0 ) ( 4 , 3 ) ( 3 , 4 ) ( 0 , 5 ) (-3 , 4 ) (-4 , 3 ) (-5 , 0 ) (-4 ,-3 ) (-3 ,-4 ) ( 0 ,-5 ) ( 3 ,-4 ) ( 4 ,-3 ) All points lie on a circle with radius 5 units and centre at the origin. x Β² + y Β² = 25 x Β² + y Β² = r Β² For any point on the circle, For any radius... NOTICE
  • 4. The Equation of a Circle with centre at the Origin NOTE SLIDE Higher Maths 2 4 Circles UNIT OUTCOME x Β² + y Β² = r Β² For any circle with radius r and centre the origin, x y The β€˜Origin’ is the point (0,0) origin Example Show that the point ( - 3 , ) lies on the circle with equation 7 x Β² + y Β² = 16 x Β² + y Β² = ( -3 ) Β² + ( ) Β² 7 = 9 + 7 = 16 Substitute point into equation: The point lies on the circle. LEARN THIS
  • 5. The Equation of a Circle with centre ( a , b ) NOTE SLIDE Higher Maths 2 4 Circles UNIT OUTCOME ( x – a ) Β² + ( y – b ) Β² = r Β² For any circle with radius r and centre at the point ( a , b ) ... Not all circles are centered at the origin. x y ( a , b ) r Example Write the equation of the circle with centre ( 3 , -5 ) and radius 2 3 . ( x – a ) Β² + ( y – b ) Β² = r Β² ( x – 3 ) Β² + ( y – ( -5 ) ) Β² = ( ) Β² 2 3 ( x – 3 ) Β² + ( y + 5 ) Β² = 12 LEARN THIS
  • 6. NOTE SLIDE Higher Maths 2 4 Circles UNIT OUTCOME The General Equation of a Circle ( x + g ) 2 + ( y + f ) 2 = r 2 ( x 2 + 2 g x + g 2 ) + ( y 2 + 2 fy + f 2 ) = r 2 x 2 + y 2 + 2 g x + 2 f y + g 2 + f 2 – r 2 = 0 x 2 + y 2 + 2 g x + 2 f y + c = 0 c = g 2 + f 2 – r 2 r 2 = g 2 + f 2 – c r = g 2 + f 2 – c Try expanding the equation of a circle with centre ( - g , - f ) . General Equation of a Circle with center ( - g , - f ) and radius r = g 2 + f 2 – c this is just a number... LEARN
  • 7. NOTE SLIDE Higher Maths 2 4 Circles UNIT OUTCOME Circles and Straight Lines A line and a circle can have two, one or no points of intersection. r A line which intersects a circle at only one point is at 90Β° to the radius and is is called a tangent . two points of intersection one point of intersection no points of intersection REMEMBER
  • 8. NOTE SLIDE Higher Maths 2 4 Circles UNIT OUTCOME Intersection of a Line and a Circle Example Find the intersection of the circle and the line 2 x – y = 0 x 2 + ( 2 x ) 2 = 45 x 2 + 4 x 2 = 45 5 x 2 = 45 x 2 = 9 x = 3 or -3 y = 2 x x 2 + y 2 = 45 Substitute into y = 2 x : How to find the points of intersection between a line and a circle: β€’ rearrange the equation of the line into the form y = m x + c β€’ substitute y = m x + c into the equation of the circle β€’ solve the quadratic for x and substitute into m x + c to find y y = 6 or -6 Points of intersection are ( 3,6 ) and ( -3,-6 ) .
  • 9. NOTE SLIDE Higher Maths 2 4 Circles UNIT OUTCOME Intersection of a Line and a Circle (continued) Example 2 Find where the line 2 x – y + 8 = 0 intersects the circle x 2 + y 2 + 4 x + 2 y – 20 = 0 x y x 2 + ( 2 x + 8 ) 2 + 4 x + 2 ( 2 x + 8 ) – 20 = 0 x 2 + 4 x 2 + 32 x + 64 + 4 x + 4 x + 16 – 20 = 0 5 x 2 + 40 x + 60 = 0 5 ( x 2 + 8 x + 12 ) = 0 5 ( x + 2 )( x + 6 ) = 0 x = -2 or -6 Substituting into y = 2 x + 8 points of intersection as ( -2,4 ) and ( -6,-4 ) . Factorise and solve
  • 10. NOTE SLIDE Higher Maths 2 4 Circles UNIT OUTCOME The Discriminant and Tangents x = - b b 2 – ( 4 ac ) Β± 2 a b 2 – ( 4 ac ) Discriminant The discriminant can be used to show that a line is a tangent: β€’ substitute into the circle equation β€’ rearrange to form a quadratic equation β€’ evaluate the discriminant y = m x + c b 2 – ( 4 ac ) > 0 Two points of intersection b 2 – ( 4 ac ) = 0 The line is a tangent b 2 – ( 4 ac ) < 0 No points of intersection r REMEMBER
  • 11. NOTE SLIDE Higher Maths 2 4 Circles UNIT OUTCOME Circles and Tangents Show that the line 3 x + y = -10 is a tangent to the circle x 2 + y 2 – 8 x + 4 y – 20 = 0 Example x 2 + (- 3 x – 10 ) 2 – 8 x + 4 (- 3 x – 10 ) – 20 = 0 x 2 + 9 x 2 + 60 x + 100 – 8 x – 12 x – 40 – 20 = 0 10 x 2 + 40 x + 40 = 0 b 2 – ( 4 ac ) = 40 2 – ( 4 Γ— 10 Γ— 40 ) = 0 = 1600 – 1600 The line is a tangent to the circle since b 2 – ( 4 ac ) = 0 x y
  • 12. NOTE SLIDE Higher Maths 2 4 Circles UNIT OUTCOME Equation of Tangents To find the equation of a tangent to a circle: β€’ Find the center of the circle and the point where the tangent intersects β€’ Calculate the gradient of the radius using the gradient formula β€’ Write down the gradient of the tangent β€’ Substitute the gradient of the tangent and the point of intersection into y – b = m ( x – a ) REMEMBER Straight Line Equation y – b = m ( x – a ) m tangent = – 1 m radius x 2 – x 1 y 2 – y 1 m radius = r