Transcript Lesson 7 - Circles

Circles, Tangents and
Chords
Objectives
•To learn 3 circle theorems
•To use these theorems to solve circle questions
Keywords
Chords, normal
© Christine Crisp
Circle Problems
Tangents to Circles
Some properties of circles may be needed in solving
problems. This is the 1st one
 The tangent to a circle is perpendicular to the
radius at its point of contact
A line which is
perpendicular to a
tangent to any curve
is called a normal.
x
radius
tangent
For a circle, the
radius is a normal.
Circle Problems
Tangents to Circles
Diagrams are very useful when solving problems
involving circles
e.g.1 Find the equation of the tangent at the point
(5, 7) on a circle with centre (2, 3)
Method: The equation of any
gradient m
straight line is y  mx  c .
x (5, 7) We need m, the gradient of
gradient m 1
the tangent.
y 2  y1
• The
Findtangent

(2, 3) x
m 1 using
to a m
circle
is
x 2  x1
tangent
perpendicular to the radius
at its point of contact
1
• Find m using m 2  
m1
• Substitute for x, y, and m in y  mx  c to find c.
Circle Problems
e.g.1 Find the equation of the tangent at the point
(5, 7) on a circle with centre (2, 3)
Solution: m1 
gradient m
gradient m1
x
y 2  y1
(5, 7)

(2, 3) x
m  m2  
tangent
x 2  x1
m1 
1
73
52


3
m  
3
4
m1
y  mx  c 
4
y3 xc
4
Substitute the point that is on the tangent, (5, 7):
 7   3 (5)  c  43  c
4

4
y   3 x  43 or
4
4
4 y  3 x  43
Circle Problems
e.g.2 The centre of a circle is at the point C (-1, 2).
The radius is 3. Find the length of the
tangents from the point P ( 3, 0).
Method: Sketch!
tangent
• Use 1 tangent and join
the radius.
The required length is AP.
C (-1, 2) x
• Find CP and use Pythagoras’
20
theorem for triangle CPA
3
Solution:
x
tangent

A
CP 

11
P (3,0)
( 3  (  1 ))
AP
2
 PC
2
2
 (0  2)
d
2
2
( x 2  x 1 )  ( y 2  y1 )
 CP  16  4 
2
 AC  AP 
20  9  11
2
20
Circle Problems
Exercises
Solutions are on the next 2 slides
1. Find the equation of the tangent at the
point A(3, -2) on the circle x 2  y 2  13
Ans:
2 y  3 x  13
2. Find the equation of the tangent at the
point A(7, 6) on the circle ( x  4) 2  ( y  2) 2  25
Ans: 3 x  4 y  45
Circle Problems
1.
Find the equation of the tangent at the point
A(3, -2) on the circle x 2  y 2  13
Solution: Centre is (0, 0). Sketch!
Gradient of radius,
m1 

y 2  y1
x 2  x1
 m1 
Gradient of tangent,
m2  
1
m1
 m
2
(0, 0)
x
gradient m 1
3
x (3, -2)
3
2
y  mx  c   2 
 Equation of tangent is
gradient
m
3
2
( 3)  c   13  c
2
y  3 x  13 or 2 y  3 x  13  0
2
2
Circle Problems
2. Find the equation of the tangent at the
point A(7, 6) on the circle ( x  4) 2  ( y  2) 2  25
Solution: Centre is (4, 2).
Gradient of radius,
m1 

y 2  y1
x 2  x1
 m1 
gradient m
gradient m 1
4
x
(7, 6)
(4 , 2) x
tangent
3
Gradient of tangent,
m2  
1
m1

 m2  
y
3
4
3
y  mx  c  6  
4
x
 45  c
45
4
or
4
3 x  4 y  45
3
4
(7)  c
Circle Problems
Chords of Circles
Another useful property of circle is the following:
 The perpendicular from the centre to a chord
bisects the chord
x
chord
Circle Problems
e.g. A circle has equation x 2  y 2  6 x  8 y  21  0
The point M (4, 3) is the mid-point of a chord.
Find the equation of this chord.
Method: We need m and c in
y  mx  c
•
x
M (4, 3) x
chord
•
•
•
Complete the square to find
the centre
Find the gradient of the
radius
Find the gradient of the
chord
Substitute the coordinates of
M into y  mx  c to find c.
Circle Problems
e.g. A circle has equation x 2  y 2  6 x  8 y  21  0
The point M (4, 3) is the mid-point of a chord.
Find the equation of this chord.
Solution:
2
2
x  y  6 x  8 y  21  0
2
2
( x  3)  9  ( y  4)  16  21  0
C (3, 4)x
 ( x  3) 2  ( y  4) 2  4
 Centre C is (3, 4)
m
m1
M (4, 3) x
chord
 y1 time: Could
4 you
3 have
Tip toy 2save
 m1 
 1
x 2 centre
 x1
got the
without 3  4
completing
the square?
1
m1 
m2  
m1
 m 1
y  mx  c  3  1(4)  c   1  c
 chord is y  x  1
Circle Problems
Exercise
1.
A circle has equation x  y  2 x  10 y  18  0
(a) Find the coordinates of the centre, C.
(b) Find the equation of the chord with midpoint (2, 6).
Solution: (a) ( x  1) 2  1  ( y  5) 2  25  18  0
2

(b)
m
C
M ( 2, 6)
m1 x
(1, 5) x
chord
2
2
2
( x  1)  ( y  5)  8
 Centre is ( 1, 5 )
m1 
m2  
y 2  y1
x 2  x1
1
m1
 m1 

65
21
1
m  1
Equation of chord is y   x  c
( 2, 6) on the chord  6  2  c  8  c
Equation of chord is y   x  8
Circle Problems
Semicircles
The 3rd property of circles that is useful is:
 The angle in a semicircle is a right angle
P
B
Q
x
A
diameter

APB  90


AQB  90

Circle Problems
e.g. A circle has diameter AB where A is ( -1, 1)
and B is (3, 3). Show that the point P (0, 0) lies
on the circle.
Method: If P lies on the
B(3, 3)
diameter
circle the lines AP and BP
will be perpendicular.
Solution: m 
x
m2
A(-1, 1) m 1
x 2  x1
Gradient of AP: m 1 
Gradient of BP:
P(0, 0)
y 2  y1
So, m1  m 2  1
m2 
1 0
1 0
30
30
Hence  APB  90 and P is on the circle.
 1
1
Circle Problems
Exercise
1.
A, B and C are the points (3, 5), ( -2, 4) and (1, 2)
respectively. Show that C lies on the circle
with diameter AB.
Solution: Gradient of AC
m
y 2  y1
x 2  x1
 m1 
25
1 3

3
2

A(3, 5)
3
diameter
2
x
Gradient of BC
m2 
24
1  ( 2 )

2
3

2
3
 2
m 1  m 2       1
2  3
3
B(-2, 4)
m1
m2
C(1, 2)
Since AC and BC are perpendicular, C lies on the
circle diameter AB.
Summary



What is the form of an equation of a circle ?
What are the three circle theorems ?
How can we tell if a line intersects a circle,
touches it or misses it completely ?