18: circles, tangents and chords © christine crisp “teach a level maths” vol. 1: as core...
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18: Circles, Tangents 18: Circles, Tangents and Chordsand Chords
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 1: AS Core Vol. 1: AS Core ModulesModules
Circle Problems
Module C1AQA Edexc
el
OCR
MEI/OCR
Module C2
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Circle Problems
x
radius
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
tangent
Tangents to Circles
A line which is perpendicular to a tangent to any curve is called a normal.For a circle, the radius is a normal.
Circle Problems
x
Diagrams are very useful when solving problems involving circlese.g.1 Find the equation of the tangent at the
point (5, 7) on a circle with centre (2, 3)
(2, 3)
(5, 7)x
tangent
The tangent to a circle is perpendicular to the radius at its point of contact
Method: The equation of any straight line is .
cmxy
1mgradient
mgradient
12
1
mm • Find m
using• Substitute for x, y, and m in to find c. cmxy
• Find using 1m12
12
xx
yym
We need m, the gradient of the tangent.
Tangents to Circles
Circle Problems
e.g.1 Find the equation of the tangent at the point (5, 7) on a circle with centre (2, 3)
cmxy
Solution:12
121 xx
yym
12
1
mmm
Substitute the point that is on the tangent, (5, 7):
x(2, 3)
(5, 7)x
tangent
1mgradient
mgradient
3
4
25
371
m
443
43 xy
c )5(743 c
443
4334 xyor
cxy 43
4
3 m
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).
xC (-1, 2)
Solution:
212
212 )()( yyxxd
P (3,0)x
3
Method: Sketch!
• Find CP and use Pythagoras’ theorem for triangle CPA
A
222 ACPCAP 11920 AP
tangent
tangent• Use 1 tangent and
join the radius.The required length is
AP.
22 )20())1(3( CP 20416 CP
20
11
Circle ProblemsExercise
s
1. Find the equation of the tangent at
the point A(3, -2) on the circle 1322 yx
2. Find the equation of the tangent at
the point A(7, 6) on the circle 25)2()4( 22 yx
Ans: 1332 xy
Solutions are on the next 2 slides
Ans: 4543 yx
Circle Problems
1. Find the equation of the tangent at the
point
A(3, -2) on the circle
1322 yx
Solution: Centre is (0, 0).
Sketch!
Equation of
tangent is
2
1323 xy o
r
01332 xy
12
121 xx
yym
Gradient of
radius, 3
21
m
Gradient of
tangent,
12
1
mm
2
3 m
cmxy c2
13c )3(223
gradient x
x(0, 0)
(3, -2)
gradient m
1m
Circle Problems
2. Find the equation of the tangent at
the point A(7, 6) on the circle 25)2()4( 22 yx
Solution: Centre is (4,
2).
445
43 xy 4543 yx o
r
Gradient of
tangent,
4
32 m
12
1
mm
3
41 m
Gradient of
radius,
12
121 xx
yym
cmxy
c445
c )7(643
gradient
(4 , 2)
(7, 6)x
tangent
mgradient
x
1m
Circle Problems
x
Chords of Circles
The perpendicular from the centre to a chord bisects the chord
chord
Another useful property of circle is the following:
Circle Problems
x
chord
The point M (4, 3) is the mid-point of a chord. Find the equation of this chord.
e.g. A circle has equation
0218622 yxyx
)3,4(M x• Find the gradient of the
radius
Method: We need m and c in cmxy
• Complete the square to find the centre
• Find the gradient of the chord
• Substitute the coordinates of M into to find c.
cmxy
Circle Problems
)3,4(M
x
chord
x
02116)4(9)3( 22 yx
)4,3(Centre C is 4)4()3( 22 yx)4,3(C
143
341
m12
121 xx
yym
1 m1
21
mm
c 1c )4(131 xy is chord
Solution:
0218622 yxyx
1mm
cmxy
The point M (4, 3) is the mid-point of a chord. Find the equation of this chord.
e.g. A circle has equation
0218622 yxyx
Tip to save time: Could you have got the centre without completing the square?
Circle Problems
(b)x
xchord
)6,2(M
Exercise1. A circle has equation
(a) Find the coordinates of the centre, C. (b) Find the equation of the chord with mid-point (2, 6).
01810222 yxyx
Solution:
(a)
01825)5(1)1( 22 yx8)5()1( 22 yx
Centre is ( 1,
5 ) )5,1(C
1m
m
1 m
Equation of chord
is
cxy )6,2( on the
chord
c 26 c 8
8 xyEquation of chord
is
12
121 xx
yym
12
1
mm
112
561
m
Circle Problems
x
Semicircles
The angle in a semicircle is a right angle
diameter
P
Q
A
B
90APB 90AQB
The 3rd property of circles that is useful is:
Circle Problems
x
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.
diameter
A(-1, 1)
B(3, 3)
Method: If P lies on the circle the lines AP and BP will be perpendicular.
Solution: 12
12
xx
yym
101
011
m
P(0, 0)
Hence and P is on the circle. 90APB
Gradient of AP:
Gradient of BP:
103
032
m
So, 121 mm
2m
1m
Circle Problems
B(-2, 4)2m
diameter
C(1, 2)
1m
A(3, 5)
x
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.
13
2
2
321
mm
12
12
xx
yym
2
3
2
3
31
521
m
3
2
3
2
)2(1
422
m
Solution:
Gradient of BC
Since AC and BC are perpendicular, C lies on the circle diameter AB.
Gradient of AC
Circle Problems
Circle Problems
The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.
Circle Problems
The tangent to a circle is perpendicular to the radius at its point of contact
The perpendicular from the centre to a chord bisects the chord
The angle in a semicircle is a right angle
Properties of Circles
Diagrams are nearly always needed when solving problems involving circles.
A line perpendicular to a tangent to any curve is called a normal. The radius of a circle is therefore a normal.
Circle Problems
e.g.1 Find the equation of the tangent at the point (5, 7) on a circle with centre (2, 3)
cmxy
Solution:12
121 xx
yym
12
1
mmm
Substitute the point that is on the tangent, (5, 7):
x(2, 3)
(5, 7)x
tangent
1mgradient
mgradient
3
4
25
371
m
4
32 m
443
43 xy
c )5(743 c
443
4334 xyor
cxy 43
Circle Problems
)3,4(M
x
chord
x
02116)4(9)3( 22 yx
)4,3(Centre C is
4)4()3( 22 yx)4,3(C
143
341
m12
121 xx
yym
1 m1
21
mm
c 1c )4(131 xy is chord
Solution:
0218622 yxyx
1mm
cmxy
The point M (4, 3) is the mid-point of a chord. Find the equation of this chord.
e.g. A circle has equation
0218622 yxyx
Circle Problems
x
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.
diameter
A(-1, 1)
B(3, 3)
Method: If P lies on the circle the lines AP and BP will be perpendicular.
Solution: 12
12
xx
yym
101
011
m
P(0, 0)
Hence and P is on the circle. 90APB
Gradient of AP:
Gradient of BP:
103
032
m
So, . 121 mm
1m
2m