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Higher Maths
Strategies
The Circle
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Maths4Scotland
Higher
The following questions are on
The Circle
Non-calculator questions will be indicated
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Maths4Scotland
Higher
Find the equation of the circle with centre
(–3, 4) and passing through the origin.
Find radius (distance formula):
You know the centre:
Write down equation:
r 5
(3, 4)
( x  3)2  ( y  4)2  25
Hint
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Maths4Scotland
Higher
Explain why the equation x2
does not represent a circle.
Consider the 2 conditions
 y2  2x  3 y  5  0
1. Coefficients of x2 and y2 must be the same.
2. Radius must be > 0
g  1,
Calculate g and f:
Evaluate
Deduction:
g  f c
2
2
f 
g 2  f 2  c  0 so
3
2
 
(1)  
2
i.e. g 2  f 2  c  0
3
2
2
5

1
4
1  2 5  0
g 2  f 2  c not real
Equation does not represent a circle
Hint
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Maths4Scotland
Higher
Find the equation of the circle which has P(–2, –1) and Q(4, 5)
as the end points of a diameter.
Q(4, 5)
C
Make a sketch
P(-2, -1)
(1, 2)
Calculate mid-point for centre:
Calculate radius CQ:
Write down equation;
r  18
 x  1   y  2   18
2
2
Hint
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Higher
Find the equation of the tangent at the point (3, 4) on the circle
x2  y 2  2x  4 y 15  0
Calculate centre of circle:
P(3, 4)
(1, 2)
Make a sketch
O(-1, 2)
Calculate gradient of OP (radius to tangent)
Gradient of tangent:
m  2
Equation of tangent:
y  2 x  10
m
1
2
Hint
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Higher
The point P(2, 3) lies on the circle ( x  1)2  ( y 1)2  13
Find the equation of the tangent at P.
Find centre of circle:
P(2, 3)
(1, 1)
Make a sketch
O(-1, 1)
Calculate gradient of radius to tangent
Gradient of tangent:
3
m
2
Equation of tangent:
2 y  3x  12
m
2
3
Hint
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Maths4Scotland
Higher
O, A and B are the centres of the three circles shown in
the diagram. The two outer circles are congruent, each
touches the smallest circle. Circle centre A has equation
 x  12 
2
  y  5   25
2
The three centres lie on a parabola whose axis of symmetry
is shown the by broken line through A.
a) i) State coordinates of A and find length of line OA.
ii) Hence find the equation of the circle with centre B.
b) The equation of the parabola can be written in the form y  px( x  q)
A is centre of small circle
A(12,  5)
Find OA (Distance formula)
Use symmetry, find B
B(24, 0)
Find radius of circle A from eqn.
Find radius of circle B
13  5  8
Eqn. of B
Points O, A, B lie on parabola
– subst. A and B in turn
0  24 p(24  q)
5  12 p(12  q)
Solve:
Find p and q.
13
5
( x  24)2  y 2  64
p
5
,
144
q  24
Hint
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Higher
Circle P has equation x2  y 2  8x 10 y  9  0 Circle Q has centre (–2, –1) and radius 22.
a) i) Show that the radius of circle P is 42
ii) Hence show that circles P and Q touch.
b) Find the equation of the tangent to circle Q at the point (–4, 1)
c) The tangent in (b) intersects circle P in two points. Find the x co-ordinates of the points of
intersection, expressing your answers in the form a  b 3
Find centre of circle P:
Find radius of circle :P:
(4, 5)
Find distance between centres
72  6 2
Gradient of radius of Q to tangent:
Equation of tangent:
m  1
Deduction:
= sum of radii, so circles touch
Gradient tangent at Q:
m 1
y  x5
2
2
Solve eqns. simultaneously x  y  8 x  10 y  9  0
y  x5
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42  52  9  32  4 2
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Soln:
22 3
Hint
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Maths4Scotland
Higher
For what range of values of k does the equation
represent a circle ?
g  2k ,
Determine g, f and c:
State condition
g  f c  0
2
2
5k 2  k  2  0
Simplify
f  k,
x2  y 2  4kx  2ky  k  2  0
c  k  2
Put in values
(2k )2  k 2  (k  2)  0
Need to see the position
of the parabola
Complete the square

1
5



5 k2  k  2

5k 

1
5 k 
10

1
10
2
2
Previous
1 
2
100 

195
100
Minimum value is
195
1
when k  
100
10
This is positive, so graph is:
Expression is positive for all k:
So equation is a circle for all values of k.
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Hint
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Maths4Scotland
Higher
For what range of values of c does the equation
represent a circle ?
Determine g, f and c:
g  3,
State condition
g2  f 2  c  0
Simplify
94c  0
Re-arrange:
x2  y 2  6 x  4 y  c  0
f  2,
c?
Put in values
32  (2)2  c  0
c  13
Hint
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Maths4Scotland
Higher
The circle shown has equation ( x  3)2  ( y  2)2  25
Find the equation of the tangent at the point (6, 2).
Calculate centre of circle:
(3,  2)
Calculate gradient of radius (to tangent)
4
m
3
3
4
Gradient of tangent:
m
Equation of tangent:
4 y  3x  26
Hint
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Maths4Scotland
Higher
When newspapers were printed by lithograph, the newsprint had
to run over three rollers, illustrated in the diagram by 3 circles.
The centres A, B and C of the three circles are collinear.
The equations of the circumferences of the outer circles are
( x  12)2  ( y  15)2  25 and ( x  24)2  ( y  12)2  100
Find the equation of the central circle.
Find centre and radius of Circle A
(12,  15)
Find centre and radius of Circle C
(24, 12)
Find diameter of circle B
45  (5  10)  30
Use proportion to find B
25
 27  15,
45
Previous
(4,  3)
r 5
25
r  10
Equation of B
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27
B
20
362  272  45
Find distance AB (distance formula)
Centre of B
(24, 12)
(-12, -15)
36
so radius of B = 15
25
 36  20 relative to C
45
 x  4    y  3  225
2
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Hint
Maths4Scotland
Higher
You have completed all 11 questions in this presentation
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Maths4Scotland
Higher
Table of exact values
sin
cos
tan
Return
30°
45°
60°

6
1
2

4

3
1
2
1
2
3
2
1
2
3
2
1
3
1
3