Maths
Leaving Cert
Higher Level –
Differentiation
Question 6 & 7 Paper 1
By Cillian Fahy
and Darron Higgins
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Paper II Co-ordinate Geometry: Question 1
Paper II Q.1 and 3................................................................................................................................... 3
Co-ordinate Geometry ............................................................................................................................ 3
The Line Q.3 ............................................................................................................................................ 3
1.
Basic questions involving the formulae given. .............................................................................. 3
Parametric equations of a line ............................................................................................................... 7
Linear Transformations ......................................................................................................................... 7
The Circle Q.1 ....................................................................................................................................... 14
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Paper II Q.1 and 3
Co-ordinate Geometry
This is a very popular combination and most students attempt one if not both of these question. If you are
comfortable with co-ordinate geometry of the line then you should do the question on the circle. The
questions asked in the circle rely heavily on the formula and techniques learnt in the line.
The Line Q.3
This question can be broken down into three sections.
1. Basic questions involving the use of the equation of the line formulae.
2. Parametric Equations
3. Linear Transformations
1.
Basic questions involving the formulae given.
One of the biggest challenges in the past was the fact that the formulae had to be learnt. Therefore,
mistakes here would lead to incorrect answers and to questions that would not work out. Now however
the formulae are all given in the table book page 18 & 19. Familiarise yourself with layout and format of
these formulae it can be different to your text book.
With that said the formula for concurrent lines is not given
If L1 : a1 x  b1 y  c1  0 and L2 : a2 x  b2 y  c2  0 are the equation of two non-parallel
lines then the equation
 (a1 x  b1 y  c1 )   (a2 x  b2 y  c2 )  0
Represents all lines concurrent with L1 and L2 .
Also note that you must learn the proof for ‘the perpendicular distance formula’ and for ‘the angle
between two lines formula’.
The secret to this topic is to think through the question, formulate a plan and make sure that you do not
jump to conclusions, such as assuming that lines are parallel or perpendicular. Remember all diagrams are
not to scale and can’t be measured as part of your answer.
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You should try to approach all questions on the line in the same manner,
1.
2.
3.
4.
5.
E.g.
Read the question carefully and identify what is being asked,
Draw a sketch of the information given, this can be a big help!
Decide what formulae to use.
Formulate a plan before you start.
Take your time and make sure that you don’t make silly mistakes
The equation of the line L is 14 x  6 y  1  0 .
{2000, Paper II Q.3 (a)}
Find the equation of the line perpendicular to L
that contains the point  3, 2  .
y  y1  m  x  x1 
Ans.
L:
……….[formula to find the equation of a line]
Note: Think about what you need for
this formula.
14 x  6 y  1  0
6 y  14 x  1
y
Need a point ….given  3, 2 
14
1
x
6
6
Therefore, slope of L is 
Therefore,
which gives
and the slope … not given!
Therefore, need to find this first.
6
3
14
. Hence, perpendicular slope is
or
14
7
6
3
 x  3
7
3x  7 y  23  0
y  (2) 
Although this is a very straightforward question the methods used to solve it are replicated when dealing
with more complicated questions, such as this one:
{2005, Paper II Q.5 (b)}
E.g.
The line K passes through the point  4, 6  and has a slope m, where m  0 .
(i)
(ii)
(iii)
Ans.
(i)
Write down the equation of K in terms of m.
Find, in terms of m, the co-ordinates of the points where K intersects the axes.
The area of the triangle formed by K, the x-axis and the y-axis is 54 square units.
Find the possible values of m.
y  6  m  x  4
mx  y  6  4m  0
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Same as above. Identify the formula
needed and apply it.
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(ii)
K: mx  y  6  4m  0
cuts x-axis when y  0
cuts y-axis when x  0
 mx  (0)  6  4m  0
 m(0)  y  6  4m  0
6  4m
m
 6  4m 

,0
 m

x
(iii)
 y  6  4m
 (0, 6  4m)
Note: This is all
the same as in
the Junior Cert.
Don’t be put off
by the m’s. Just
remember to do
what you have
always done.
Given that the area of the triangle is 54 units squared.
Note: a quick sketch of the question
always helps. Now we can see that we
(0, 6  4m)
have the 3 vertices and one is  0, 0  .
 6  4m 
,0

 m

Area 

1
x1 y2  x2 y1
2
1
6  4m 
 0  0   
  6  4m   54
2
 m 
 16m2  6  m  36  0
3
 m  or m  3
4
Note: apply the formula and methods as you would any question. Sketch a picture to get a better idea. If
asked to find two lines simply focus on finding one and you will find that the other one will turn up due to
a quadratic equation, modulus brackets of a square root.
{1999, Paper II Q.3 (c)}
E.g.
A line containing the point  4, 2  has a slope of m, where m  0 .
This line intercepts the x-axis at  x1 , 0  and the y-axis at  0, y1  .
Given that x1  y1  3 , find the slopes of the two lines the satisfy this condition.
Find the measure of the acute angle between these two lines and give your answer to the nearest
degree.
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Ans.
Equation of the line is y  2  m  x  4 
Cuts the x-axis when y  0
 x1 
2  4m
m
Cuts the y-axis when x  0
 y1  4m  2
Given that x1  y1  3 so
2  4m
 4m  2  3
m
 4m2  9m  2  0
Note: as with the last two examples
find the equation using Junior Cert
methods, don’t worry about the
second line for the moment it will turn
up.
Find the points as above.
Remember all information in the
question must be used so substitute
values into the given equation.

  m  2  4m  1  0
1
 m  2 or m 
4
Now we have two values for m as
asked!
Don’t forget to find the second part. The measure
of the acute angle between the lines.
tan  
m1  m2
1  m1m2
tan  
2  14
1  (2)( 14 )
tan  
7
6
  49
Here we have dealt with some of the main areas that come up in the co-ordinate geometry of the line.
However, you should make sure that you are able to use all of the other formula given on page 18 of your
table book. Perpendicular distance between a point and a line is very common as is dividing a line in a
given ratio. As with the examples above simply identify the formula required, substitute in the values
given and draw a quick sketch if you are still unsure of what to do.
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Parametric equations of a line
A parametric equation represents a line using a single variable (usually t). This is not a common question
and last came up in 1998. However, it is important to know and can also come up in the Q.1 The Circle.
{1998, Paper II Q.3 (a)}
The parametric equations x  3  4t and y  1  2t represent a line, where t  R .
Find the Cartesian equation of the line.
Ans.
x  3  4t
3 x
t 
4
Therefore,
giving
and
y  1  2t
y 1
2
t 
3  x y 1

4
2
x  2y 5  0
Note: Re-write each equation in terms
of t. Therefore we can let them equal
and write an equation in terms of x
and y.
Linear Transformations
Another part to this question is Linear Transformations and this regularly comes up as a part (b) or (c) in
Q.3.
A linear transformation is a transformation of the form
f  x, y    ax  by, cx  dy 

This is often written as : f  x, y   x , y
where
'
'

x '  ax  by
y '  cx  dy
Note: There is a separate rule for
moving the x and the y coordinates
It is important to know and understand the properties of Linear Transformations as these can be used in
your answer.
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N.B.
(i)
The origin  0, 0  is mapped onto the origin.
(ii)
(iii)
(iv)
(v)
A line is mapped onto a line.
A line segment is mapped onto a line segment.
Pairs of parallel lines are mapped onto pairs of parallel lines.
Parallelograms are mapped onto parallelograms.
(i)
In general pairs of perpendicular lines are not mapped onto pairs of
perpendicular lines.
In general distances and areas are not preserved.
(ii)
The secret to this question is to simply do what you are told. In the example below we will cover the main
areas of Linear Transformations. If you understand the notation used then it will be clear to see that if you
take each section step by step it is very easy to get full marks.
a  2, 2  , b  4,1 , c  2, 1 , d  2,5 are four points. A transformation f is
Eg.
given by
f  x, y    x' , y ' 
where
x'  x  2 y
y'   x  y
f  a  , f b , f c  , f  d 
(a)
Find
(b)
Show that
(i)
(ii)
(iii)
(c)
ab  f  a  f  b 
area of abc  1
ab
cd

3
f  a  f b
area of
f  a  f b  f  c 
f c f d 
p is the midpoint of [bc] , q is the midpoint of  f  b  f  c  
g is the centroid of abc , h is the centroid of
f  a  f b  f  c 
Investigate if
(d)
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(i)
f  p  q
(ii)
f g  h
s  7,8 is the image of r  x, y  under f. Find the coordinates of r.
Maths LC HL Co-ordinate Geometry © mocks.ie
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Ans:
(a)
(b)
Note: There is a separate rule for
moving the x and y values.
 x, y    x  2 y ,  x  y 
a(2, 2)  (2  2  2  , 2  2)   2, 4 
 f a
b  4,1   4  2 1 , 4  1   6,3 
 f b 
c  2, 1   2  2  1 ,   2   1   4,1
 f c
d  2,5   (2  2  5  , 2  5)  12,3
 f d 
(i)
ab  f  a  f  b 
 4  2
13 
2
 1  2 
2

6  2
2
  3  4 
2
65
 ab  f  a  f  b  as asked.
(ii)
area of abc  1
3
area of
f  a  f b  f  c 
a  2, 2    0, 0 
f  a   (2, 4)  (0, 0)
b  4,1   2,3
f (b)   6, 3   8,1
c  2, 1   4,1
f  c    4,1   2,5 
 area  1 2(1)  (4)(3)
2
7
 area abc 
Note: As said earlier, simply do
what you are told. Find the
length of ab and f (a) to f (b) and
show that they are not the same
Note: as above but
using the area of a
triangle formula
 area  1 8  5    2 1
2
 21
1 area f  a  f  b  f  c 
3
as 7  1 (21)
3
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(iii)
ab
cd

f  a  f b
Note: Again don’t get confused. Use
the appropriate formula and simply
work through the question
f c f d 
ab  13 from (i)
cd 

ab
cd
 2  2

13
52
2
  5  1  52
2
1 1

4 2

f  a  f  b   65
f c f d  

f  a  f b 
f c f d 
Therefore,
(c)
(i)
ab
cd
from (i)
12  4   (3  1)
2


65
260

2
 260
1 1

4 2
f  a  f b
f c f d 
b  4,1 c  2, 1
p is the midpoint of bc  p 1, 0 
 f  p   1  2  0  , 1  0   1, 1
q is the midpoint of [ f  b  f  c ]  q 1, 1
 f  p  q
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(ii)
a  2, 2  , b  4,1 , c  2, 1
Note: the centroid does not come up very often
and can be overlooked.
g is the centroid of abc . Thus
A triangle with vertices
a  x1 , y1  , b  x2 , y2  , c  x3 , y3 
 2  4  2 2  1  1 
g 
,

 x  x2  x3 y1  y2  y3 
3
3


,
has centroid g   1

3
3


4 2

,



h is
 3 3  the centroid of f  a  f  b  f  c  . Thus
 2  6  4 4  3  1 
h
,

3
3


  0, 2 

f g  f 4 , 2
3
3




 4  2  2 , 4  2
3
3
3
3

  0, 2 
Therefore, f  g   h as asked.
(d)
Given
 x , y    7,8
'
'
  x  2 y,  x  y    7,8 
 x  2 y  7 and  x  y  8
Solving simultaneously gives x  3 and y  5
Therefore, r   3,5
The other common type of question is to find the image of a line under a linear transformation, for
example, 2001 Paper II Question 3
Q.
f is the transformation  x, y    x ', y ' where
x '  5x  6 y
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y '  4x  3 y
L is the line x  9 y  2 .
Find the equation of f  L  where k  Z .
(i)
x '  5x  6 y
Ans.
 2 
y '  4x  3 y
x '  5x  6 y
2 y '  8x  6 y
Note: Using a simultaneous equation
re-write the two equations in terms of
x and y.
x ' 2 y '  3x
x 
x ' 2 y '
3
Putting this back into y '  4 x  3 y we get
 x ' 2 y ' 
y '  4
  3y
 3 
3 y '  4  x ' 2 y '  9 y '
3 y '  4 x ' 8 y ' 9 y
y 
4 x ' 5 y '
9
Therefore, if L: x  9 y  2
Then, f  L  :
 x ' 2 y '   4 x ' 5 y ' 

  9
2
9
 3  

x ' 2 y ' 12 x ' 15 y '  6
Note: Now we know the image of
each x and y value simply replace the
x and y values in the equation to find
the image of the line.
13x ' 17 y '  6
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Note
The following does not come up much anymore but you should know the meanings
of the words just in case.
Centroid (g): A median of a triangle is a line segment from a vertex to the midpoint of the
opposite side. The three medians of a triangle meet at a point called the centroid, g.
 x  x2  x3 y1  y2  y3 
g  1
,

3
3


g
Note: g divides each median in the
ratio 2:1. This can be particularly
important for the Vectors question.
Circumcentre (c): The Circumcentre of a triangle is the point of intersection of the
perpendicular bisectors of the three sides.
Orthocentre (h): The Orthocentre is the point of intersection of the perpendicular lines from the
vertices to the opposite sides.
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The Circle Q.1
A lot of the Circle is dependent on the Line. You will find that in many questions after two or three lines
in to the question it becomes simply a question of working through the co-ordinate geometry of the line
formulae.
The circle can be broken up into four main sections:
1. Finding the equation of a circle.
2. Finding the tangent to a circle.
3. Touching circles and intersection between circles and lines
4. Parametric equations of a circle.
1.
Finding the equation of a circle.
If given a circle centre and radius you have the choice of three formula to use and this depends on
where the centre of the circle is.
(i)
Circles with Centre (0,0)
The equation of a circle, centre  0,0  and radius length r is
x2  y 2  r 2
(ii)
General equation of a circle with centre  h, k  and radius r.
The equation of a circle, centre  h, k  and radius length r is
 x  h   y  k 
2
2
 r2
Note: Only the second equation is given in the table book. You should know the first one.
(iii)
The general equation for a circle is given by
x2  y 2  2 gx  2 fy  c  0
This represents a circle with
(i) Centre    g ,  f 
 ii  Radius 
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Note: to find the centre, find  12 the xcoefficient and  12 the y-coefficient
g2  f 2  c
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Page 14
The following 4 examples cover the basics of the circle.
E.g. 1 Find the equation of each of the following circles with centre  0, 0 
 i  S1 , the circle which contains the point 1,4 
 ii  S2 , the circle which has the line x  3 y  10  0 as a tangent.
Solution:
 i  Let the eqt. of S1 be x 2  y 2  r 2 (where r is unknown)
Since 1,4  in on the circle we know that
2
2
x 2  y 2  r 2  1   4   r 2
Note: Centre is  0, 0  so use the first
formula.
 1  16  r 2
 r 2  17
 the eqt of S1 is x 2  y 2  17
 ii  Given that x  3 y  10  0 is a tangent to the circle we know that the distance from
this line to the centre  0,0  is the same as the radius.

ax1  by1  c
 distance 
a 2  b2

r 




1 0   3  0   10
12   3
2
10
10
 10
Therefore the equation of the circle is,
x2  y 2 
 10 
2
Note: If you draw a quick sketch you
will find that you are back to coordinate geometry of the line using the
perpendicular distance formula.
x 2  y 2  10
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E.g. 2 Find the equation of a circle which has the line segment from
 1,5  to  5, 3 as a diametre.
Solution: We need to find the radius and the centre of the circle before we can apply the formula.
Centre = the midpoint between  1,5  and  5, 3 .
 1  5 5  3 
 Centre  
,
   2,1
2 
 2
Radius = distance from the centre to one of the points on the circle.
 Radius 
 2  1  1  5 
2
2
 9  16
5
Therefore, the equation of the circle is
 x  2    y  1
2
2
Note: Centre is not  0, 0  so must use
2
the formula  x  h    y  k   r
2
 25
 x 2  4 x  4  y 2  2 y  1  25
This is the eqt of the circle.
 x 2  y 2  4 x  2 y  20  0
Obviously if given the equation of a circle, in the form above, and asked to find the centre and the radius,
a quick comparison with the formula will give the answers.
E.g. 3 Find the centre and the radius of the circle  x  1   y  3  16.
2
2
Solution: Comparing  x  1   y  3  16 to  x  h    y  k   r 2
2
2
2
2
We see that  h  1 ,  k  3 and r 2  16
 h  1, k  3, and r  4
 Centre   1,3 Radius  4
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Page 16
2
The g, f and c formula is used the most. It can be used in two ways. Firstly, to find the centre and radius
of a circle when given its equation. Secondly, and more importantly to find the equation of a circle when
given three pieces of information, this is a much more common question.
E.g. 4 Find the centre and the radius of the circle x 2  y 2  2 x  4 y  11  0
Solution: Comparing x 2  y 2  2 x  4 y  11  0
with x 2  y 2  2 gx  2 fy  11  0
We see that 2 g  2, 2 f  4, and c  11
 g  1, f  2, and c  11
Centre    g ,  f   1, 2 
Radius  g 2  f 2  c  1  4  11
Note: In practice it is easier to use the quick rule,
centre
1
 1

   coefficient of x,  coefficient of y 
2
2


 16
4
To find the equation of a circle given three pieces of information can be asked in three main ways.
1.
2.
3.
Given three points on a circle.
Given two points on the circle and a line containing the centre.
Given a tangent to a circle (either a line or an axis) and points on the circle.
The following three examples cover these possible questions.
1. Given three points on a circle.
E.g. 5 The points a  2, 4  , b  0, 10  and c  6, 2  are three points.
Find the equation of the circle that passes through these three points
Ans.
a  2, 4 
 2   4
2
2
 2 g  2   2 f  4   c  0
4 g  8 f  c  20
b  0, 10 
 0  10
2
2
 2 g  0   2 f 10   c  0
20 f  c  100
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…(1)
…(2)
Maths LC HL Co-ordinate Geometry © mocks.ie
Note: Substitute each of the
points into the general
equation of a circle.
This will leave you with three
equations in terms of g, f and
c. Now solve using
simultaneous Page
equations
17 as in
Algebra.
 6   2
c  6, 2 
2
2
 2 g  6   2 f  2   c  0
12 g  4 f  c  40
4 g  8 f  c  20
20 f  c  100
12 g  4 f  c  40
…(3)
…(1)
…(2)
…(3)
Solving these we get g  1 , f  3 and c  25
Note: This question is taken
from 2002 Paper II Q.1 (b) (ii).
It can be solved in a few
different ways but the method,
although slightly longer, will
always work.
Putting these back into the general equation gives us
x2  y 2  2 x  6 y  40  0
2. Given two points on the circle and a line containing the centre.
E.g. 6 A circle of radius length
20 contains the point  1,3 . Its centre lies on the line x  y  0 .
Find the equation of the two circles that satisfy these conditions.
Ans.
The centre of a circle is   g ,  f
 and this lies on
x y 0
Therefore,
or
g  f  0
g f
…(1)
20 , therefore
The radius is
g 2  f 2  c  20
or
g 2  f 2  c  20
…(2)
 1,3 is a point on the circle, therefore
 1  3
2
2
 2 g  1  2 f  3  c  0
2 g  6 f  c  10
…(3)
As above we now have three equations in terms of g, f and c
g f
Mocks.ie
…(1)
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Page 18
g 2  f 2  c  20
…(2)
2 g  6 f  c  10
…(3)
Substituting (1) into (3) we get,
2 g  6 g  c  10
c  8g  10
…(4)
Substituting (4) and (1) into (2) we get,
g 2    g    8g  10   20
2
g 2  4g  5  0
 g  1 g  5  0
 g  1
and
g 5
 f 1
 f  5
 c  18
 c  30
Therefore, the equations are
x2  y 2  2 x  2 y  18  0 and x2  y 2  10 x  10 y  30  0
Note: There are many ways to do this question. An alternative method is to find the centre and radius
using co-ordinate geometry of the line methods. From equation (1) above we know that the centre of the
circle must be   g , g  and that the distance from the centre to the point  1,3 is
20 . Using the
distance formula we could quickly find the coordinates of the centre and hence the two equations.
3. Circles touching the axis
The final example deals with circle that either touch or intersect an axis. Here it is important to sketch a
diagram of the question.
For a circle touching the x-axis
For a circle touching the y-axis
 g  r
f r
  f  g2  f 2  c
 f 2  g2  f 2  c
 g  g 2  f 2  c
r g
 g2  g2  f 2  c
  g ,  f g ,  f 
 g2  c
Mocks.ie
 f2 c
Maths LC HL Co-ordinate Geometry © mocks.ie
r f
 g,  f 
 g,  f 
Page 19
{2004 Paper II Q. 1 (c) (ii)}
E.g. 7 Find the equations of the circles that pass through the points  3, 6  and  6,3 and have the yaxis as a tangent.
 3, 6  is on the circle, therefore
Ans.
  3   6   2 g  3  2 f  6   c  0
2
Note: As before the question is
reduced down to three pieces of
information and from this three
equations can be found.
2
 6 g  12 f  c  45
…(1)
 6,3 is on the circle, therefore
  6    3  2 g  6   2 f  3  c  0
2
2
 12 g  6 f  c  45
…(2)
The y-axis is a tangent, therefore
 f2 c
…(3)
As in the last two examples we now have three equations in terms of g, f and c.
 6 g  12 f  c  45
 12 g  6 f  c  45
…(1)
 f c
…(3)
Again solving simultaneously we get
g  3 , f  3 and c  9
…(2)
2
and
g  15 , f  15 and c  225
x2  y 2  6 x  6 y  9  0
Giving the equations as
x2  y 2  30 x  30 y  225  0
and
Note: if a circle is touching a line remember that the perpendicular distance from the centre of the
circle to the line is equal to the radius.
radius =  distance 
 g,  f 
r
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Maths LC HL Co-ordinate Geometry © mocks.ie
ax1  by1  c
a 2  b2
Page 20
2. Finding the tangent to a circle.
E.g.
Find the equations of the tangents to the circle x2  y 2  12 x  10 y  51  0 from the
point 1, 0  .
Firstly draw a sketch of the question.
Notice that the perpendicular distance for
the centre (which we can find) is equal to
the radius (which we can also find).
Ans.
10
mx  y  m  0
 6,5
As always just focus on finding one
tangent. The other one will turn up.
Now find the equation of the line in terms
of m.
y  y1  m  x  x1 
y  0  m  x  1
 mx  y  m  0
Now we know that the  from mx  y  m  0 is

m  6    1 5   m
m2   1
2
10
 10
5m  5  10 m2  1
 5m  5
2
 10m2  10
3m2  10m  3  0
 3m 1 m  3  0
m 
1
or m  3
3
Therefore, the two lines are x  3 y  1  0 and 3x  y  3  0 .
Mocks.ie
Maths LC HL Co-ordinate Geometry © mocks.ie
Page 21
3. Touching circles and intersection between circles and lines
Circles can touch either externally or internally
For external we know that the distance between their centres is equal to the sum of their radii.
r1
r2
Therefore, d  r1  r2
d
For internal we know that that the difference between their centres is equal to the distance
between their centres.

r1

r2
Therefore, d  r1  r2
d
Circles that intersect create a common chord. This can be found by subtracting one equation of
the circle away from the other.
E.g.
Find the common chord to the circles
C1 : x2  y 2  2 x  4 y  1  0 and C2 : x 2  y 2  4 x  2 y  1  0
Ans.
Mocks.ie
C1 : x2  y 2  2 x  4 y  1  0
Maths LC HL Co-ordinate Geometry © mocks.ie
Page 22
C2 : x 2  y 2  4 x  2 y  1  0
2 x  6 y  0
Therefore, the common chord is x  3 y  0
Subtracting we get
4. Parametric equation of a circle.
There are two types of parametric equation used in the circle, (i) using algebra in which the letter
t is used to describe the x and y values and (ii) using trigonometry in which the angle  is used.
(i)
Using Algebra
{2003. Paper II Q.1 (a)}
 3  3t 2 6t 
For all value of t  R , the point 
lies on the circle x 2  y 2  r 2 .
,
2
2 
 1 t 1 t 
Find r, the radius of the circle.
Ans.
3  3t 2
x
1 t2
x 1  t
2
  3  3t
and
2
x  xt 2  3  3t 2
6t
1 t2
y
 6t 
y 
2 
 1 t 
2
2
y2 
xt 2  3t 2  3  x
3 x
…(1)
t 2 
x3
Now we can substitute (1) into (2) and get
 3 x 
36 

3 x 

2
y 
2
 3 x 
1 

 3 x 
36  3  x  3  x 
y2 
36
y 2  9  x2
36t 2
1  t 
2 2
…(2)
Note: By writing t in terms of x we
can substitute this into our second
equation. This gives us an equation in
terms of x and y only. Allowing us to
reform the Cartesian equation and
hence find the radius.
x2  y 2  9
Therefore, the radius is 3.
Mocks.ie
Maths LC HL Co-ordinate Geometry © mocks.ie
Page 23
(ii)
Using Trigonometry [more likely to come up]
{2002, Paper II Q.1 (a)}
The following parametric equations define a circle:
x  4  3cos , y  2  3sin  where   R .
What is the Cartesian equation of the circle?
Ans:
x  4  3cos
 cos  
and
x4
3
y  2  3sin 
sin  
y2
3
Using the trigonometric identity cos2   sin 2   1
We get,
 x4  y2

 
 1
 3   3 
2
 x  4   y  2
2
Mocks.ie
2
2
Note: Firstly write each
equation in terms of
cos  and sin  .
Then substitute these values
into cos2   sin 2   1 to
get the equation of the
circle.
9
Maths LC HL Co-ordinate Geometry © mocks.ie
Page 24