Week 5 – SS2 Second Term Further Mathematics Notes

 WEEK FIVE
TOPIC :THE CIRCLE: DEFITION , GENERAL EQUATION, EQUATION OF TANGENT TO CIRCLE AND LENGTH OF THE TANGENT
Definition:
    A circle is defines as the locus of point equidistant from a fixed point. A circle is completely specified by the centre and the radius.
    Equation of a circle with centre (a,b) and radius r.

PR = x – a, PQ=r
QR = y – b
Since PQR is the right angle triangle, we have:
PQ² = PR² + QR²
r² = (x – a)² + (y – b)²
hence, the equation of a circle with centre (a,b) and radius r is
r² = (x – a)² + (y – b)²
if the centre of the circle is the origin (0,0), the equation become x² + y² = r²

 GENERAL EQUATION OF A CIRCLE
From (x – a)² + (y – b)² = r²
r² – 2ax + a² + y² – 2by + b² – r² = 0
x² + y² – 2ax – 2by + a² + b² – r² = 0
The above equation can be written as x² + y² + 2gx + 2fy + c = 0
Where a = – g, b = -f, c = – a² + b² – r²
Hence: x² + y² + 2gx + 2fy + c = 0 is called the general equation of a circle. observe the following about the general equation

1. It is a second degree equation in x and y

   iiThe co-efficient of x² and y² are equal
   iiiIt has no xy term
   Examples:
   

2. Find the equation of a circle of centre (3, -2) radius 4 unit

   Solution:
   a = 3, b = -2 and r = 2
   (x-a)² + (y-b)² = r²
   (x-3)² + (y+2)² = 4²
   x² – 6x + 9 + y² + 4y + 4 = 16
   x² + y² – 6x + 4y + 9 + 4 – 16 = 0
   x² + y² – 6x + 4y – 3 = 0
   Find the centre and radius of a circle whose equation is x² + y² – 6x + 4y – 3 = 0

    Solution:
   x² + y² – 6x + 4y – 3 = 0
   x² – 6x + y² + 4y = + 3
   Complete the square for x and y
   x² – 6x + 9 + y² 4y + 4 = 3 + 9 + 4
   (x – 3)² + (y + 2)² = 16
   Compare with (x – a)² + (y – b) = r²
   
    Equation of a circle passing through 3 points
   Find the equation of the circumcircle of the triangle whose vertices are A (2,3) B (5,4) and C (3,7)
   Solution:
   The equation of the circle x² + y² + 2gx + 2fy + c = 0
   2² + 3² + 4g + 6f + c = 0
   5² + 4² + 10g + 8f + c = 0
   3² + 7² + 6f + 14f + c = 0
   Simplify the 3 equations
   f = – 107 / 22
   g = – 67 / 22
   c = – 312 / 11
   Hence, the equation of the circle is
   x² + y² + 2 x + 2 y + = 0
   11x² + 11y² – 67x – 107y + 312 = 0
   a = 3, b = -2, r² = 16,
   r = = 4
   hence the centre is (3, -2) and the radius is 4 unit
   Evaluation:
   

3. Find the equation of the circle (-1 -1) and radius 3
4. Find the centre and radius of the circle x² + y² – 6x + 14y + 49 = 0

    

5. EQUATION OF TANGENT TO A CIRCLE AT PONIT (x₁, y₁)
   

   
   At x₁, y₁
   x₁² + y₁² + 2gx₁ + 2fy₁ + c = 0
   C = – (x₁² + y₁² + 2gx₁ + 2fy₁) ……………… (i)
   Differentiating the equation of circle above
   2x + 2y dy / dx + 2g + 2f dy / dx = o
   Divide through bu 2
   x + y dy/dx + g + f dy/dx = 0
   (y + f) dy/dx = – (x+g) =
   The equation of the tangent at x₁, y₁
   =
   (y – y₁) (y + f) = – (x – x₁) (x + g)
   yy₁ + yf – y₁² – y₁f = – (xx₁ + xg – x₁² – x₁g)
   yy₁ + yf – y₁² – y₁f = – (xx₁ + xg + x₁² + x₁g)
   yy₁ + yf + xx₁ + xg = x₁² + x₁g + y₁² + y₁f)
   yy₁ + + xx₁ + yf + xg = x² + y² + x₁g + y₁f)
   Adding gx₁ + y₁f to both sides
   yy₁ + xx₁ + y₁f + yf + xg + gx₁ = x₁² + y₁² + x₁g + x₁g + y₁f + y₁f
   yy₁ + xx₁ + ( y₁ + y) f + (x + x₁) g = x₁² + y₁² + 2x₁g + 2y₁f
   but x₁² + y₁² + 2x₁g + 2y₁f = -C
   yy₁ + xx₁ + ( y₁ + y) f + (x + x₁) g + C = 0
   Hence the equation of the tangent to the circle x² + y² + 2gx + 2fy + C = 0 at (x₁, y₁) on the circle is xx₁ + yy₁ + (x + x₁)g + (y + y₁)f + c = 0
   Examples:
   Show that the point (2,3) lies on the circle x² + y² – 3x + 4y – 19 = 0. Hence or otherwise, determine the equation of the tangent to the circle at the point (2,3)
   Solution:
   x² + y² – 3x + 4y – 19 = 0
   At (2,3)
   2² + 3² – 3(2) + 4(3) – 19 = 0
   4 + 9 – 6 + 12 _ 19 = 0
   R.H.S = L.H.S, hence the point (2,3) lies on the circle
   x² + y² – 3x + 4y – 19 = 0
   Compare with:
   x² + y² + 2gx + 2fy + c = 0
   2g = -3, 2f = 4
   g = f = = 2 c = -19
   Equation of Tangent:
   yy₁ + xx₁ + (x + x₁) g + (y + y₁) f + c = 0
   3y + 2x + (x + 2) ( ) + (y + 3)2 – 19 = 0
   3y + 2x – ( ) – 3 + 2y + 6 – 19 = 0
   6y + 4x – 3x -6 + 4y + 12 – 38 = 0
   10y + x – 32 = 0
   Alternatively:
   x² + y² – 3x + 4y – 19 = 0
   2x + 2y dy/dx – 3 + dy/dx = 0
   (2y + 4) = 3 – 2x
   =
   At 2,3

   y – y₁ = m(x – x1)
   y – 3 = -1/10 (x – 2)
   10(y – 3) = -1 (x – 2)
   10y – 30 = – (x + 2)
   10y + x – 30 – 2 = 0
   10y + x – 32 = 0

    
    Evaluation
   Find the equation of the tangent to the circle

   1. x² + y² + 4x – 10y – 12 = 0 at (3,1)
   2. x² + y² – 6x – 3y = 16 at (-2,0)

       GENERAL/REVISION EVALUATION
      

      1. Find the equation of the circle with center (1,3) and radius
      2. Find the equation of the circle that passed through the point (0,0), (2,0) and (3, -1)
      3. Find the equation of the circumcircle of the triangle whose vertices are A( 2,3) B ( 5,4) and C(3,7)
      4. Find the length of the tangent to the circle x² + y² -2x -4y -4 =0 from the point ( 8, 10)

       READING ASSIGNMENT
      Read equation of a circle, Further Mathematics Project II, page 205 – 210

       WEEKEND ASSIGNMENT
      

      1. What is the radius of the circle whose equation is x² + y² – 6x – 7=0 (a) 2 (b) 3 (c) 4 (d) 9
      2. Which of the following is not an equation of a circle? (a) x² + y² =4 (b) x² + y² – 2x – 3=0 (c) x² + y² – 2xy + 4x – 6y + 1 = 0 (d) 2x² + 2y² – 6x + 4y + 3 = 0
      3. The equation of a circle with centre (-2, 5) and radius 3 units is (a) x² + y² + 4x – 10y + 20 = 0 (b) x² + y² + 4x – 10y + 26 = 0 (c) x² + y² + 4x – 10y – 38 = 0 (d) x² + y² + 4x – 10y + 39 = 0
      4. Find the coordinates of the centre of the circle 2x² + 2y² – 4x + 12y – 7 = 0 is (a) (-1, 3) (b) 1, 3) (c) (2, -6) (d) 41 (e) 10
      5. The equation of a circle of radius 3 is x² + y² + 10x – 8y + k = 0. Find the value of the constant K (a) -50 (b) 18 (c) 32 (d) 41 (e) 10

       THEORY
      

   3. The equation of a circle is x² + y² – 10x + 8y = 0 find (i) its radius (ii) its area
   4. A circle passes through the points (0,3) and (4,1), if the centre of the circle is on the x – axis, find the equation of the circle.