Slides: C2 - Chapter 4 - Coordinate Geometry

Report
C2: Coordinate geometry
Dr J Frost ([email protected])
Last modified: 14th September 2013
Midpoint of two points and distance between them
Midpoint
Distance
(1,1) (2,2)
(1.5,? 1.5)
√2
?
(1,2) (4,6)
? 4)
(2.5,
5?
(1,-4) (-3,5)
(-1,?0.5)
√97
?
(-1,-2) (-2,10)
? 4)
(-1.5,
?
√145
(a,a-1) (b,a+1)
?
([a+b]/2,
a)
√[(b-a)? 2 +4]
Centres of circles
The line PQ is a diameter of the circle centred at (2, -2). Given P is
(8, -5), find the coordinate of Q.
Q = (-4, 1) ?
The points M(3, p) and N(q, 4) lie on the circle centre (5, 6). The
line MN is a diameter of the circle. Find the value of p and q.
p = 8, q? = 7
Radii of circles
The line PQ is a diameter of a circle, where P and Q are (-1, 3) and (6, -3)
respectively. Find the radius of the circle.
r = 0.5√85
?
The points (-3, 19), (-15, 1) and (9, 1) are points on the circumference of a
circle. Show that (-3, 6) is the centre of the circle.
We just need to show that each point is equidistant from the centre.
? centre turns out to be 13.
The distance for all three points to the
Lines through circles
The line AB is a diameter of the circle centre C, where A and B are (-1, 4) and
(5,2) respectively. The line l passes through C and is perpendicular to AB. Find
the equation of l.
y = 3x -? 3
The line PQ is a chord of the circle centre (-3, 5), where P and Q are (5, 4) and
(1, 12) respectively. The line l is perpendicular to PQ and bisects it. Prove that l
passes through the centre of the circle.
Equation of perpendicular line:  =
?
When  = 3,  = −5
1

2
+
13
2
Lines through circles
The perpendicular from the
centre of a circle to a chord
bisects the chord.
We say that the radius is
the perpendicular bisector
of the chord.
Lines through circles
y
D
x
The lines  and  are chords
of a circle. The line  = 3 − 11
is the perpendicular bisector of
. The line  = − − 1 is the
perpendicular bisector of .
Find the coordinates of the centre
of the circle.
C
 
, ?−
 
A
B
Proving a triangle has a right angle
Suppose we know the three sides of a
triangle. How do we prove it has a right
angle?
5
3
It satisfies Pythagoras’ Theorem.
?
4
C
How therefore could we prove that  is
the diameter of the circle?
A
B
If AC2 + BC2 = AB2, then angle ACB is a
right angle. Thus (from GCSE circle
theorems!) AB must?be the diameter of
the circle.
Proving a triangle has a right angle
If PR is the
diameter, how
would you prove
that  = 13?
PQ2 = 208
QR2 = (a-9)2 + 36
PR2 = (a+3)2 + 4
?
Setting PQ2 + QR2 = PR2 and
solving, we find that a = 13.
Questions
1
Ex4A Q10) The points V(-4, 2a) and W(3b,
-4) lie on the circle centre (b, 2a). The line
VW is a diameter of the circle. Find the
value of a and b.
3
a = -2, b ?
=4
2
Ex4B Q1) The line FG is a diameter of the
circle centre C, where F and G are (-2, 5)
and (2, 9) respectively. The line l passes
through C and is perpendicular to FG.
Find the equation of l.
y= -x + 7 ?
4
Ex4B Q10) Find the centre of a circle with
points on its circumference of (-3,19),
(9,11) and (-15, 1).
(-3, 6)
?
Ex4B Q9) The points P(3, 16), Q(11, 12) and
R(-7, 6) lie on the circumference of a circle.
a) Find the equation of the perpendicular
bisector of:
i. PQ
ii. PR
b) Hence, find the coordinates of the
centre of the circle.
a) i) y = 2x ii) y =?-x + 9
5
b) (3, 6)
Ex4C Q9) The points A(2, 6), B(5,7) and C(8,2) lie on a circle.
a) Show that triangle ABC has a right angle.
b) Find the area of the triangle.
c) Find the centre of the circle.
?
?
?
a) AB = √10, AC = 10. BC = r90. AB2 + BC2 = AC2.
b) 15
c) (5, 2)
Equation of a circle
y
What is the equation of a
circle with radius r, centred at
the origin?
(x,y)
r
r
x
(Hint: draw a right-angled triangle inside
your circle, with one vertex at the origin and
another at the circumference)
y
r
x
x2 + ?y2 = r2
Equation of a circle
y
Now suppose we shift the
circle so it’s now centred at
(a,b). What’s the equation
now?
r
(Hint: What would the sides of this rightangled triangle be now?)
(a,b)
x
? 2 = r2
(x-a)2 + (y-b)
Equation of a circle
Centre
Radius
Equation
(0,0)
5
x2 + y?2 = 25
(1,2)
6
(x-1)2 + (y-2)
? 2 = 36
(-3,5)
?
?1
?7
?4
√3
?
2√2
?
(x+3)2 + (y-5)2 = 1
(-5,2)
?
(-6,-7)
?
(1,-1)
?
(-2,3)
?
(x+5)2 + (y-2)2 = 49
(x+6)2 + (y+7)2 = 16
(x-1)2 + (y+1)2 = 3
(x+2)2 + (y-3)2 = 8
Equation of a circle
? Show that the circle (x-3)2 + (y+4)2 = 20 passes through (5, -8).
Just substitute x=5 and y=-8 and show
? the equation holds!
? Where does the circle (x – 1)2 + (y – 3)2 = 45 meets the x-axis?
On x-axis, y = 0. So (x – 1)2 + 9 = 45.?Solving gives (7,0) and (-5,0)
? Given that AB is the diameter of a circle where A and B are (4,7) and (-8,3)
respectively, find the equation of the circle.
(x+2)2 + (y-5)2 = 400
?
? The line 4x – 3y – 40 = 0 is a tangent to the circle (x – 2)2 + (y – 6)2 =
100 at the point P(10,0). Show that the radius at P is perpendicular
to the line (as we would expect!)
Equation of line is y = 4/3 x – 40/3, so gradient is 4/3
Centre of circle is (2,6)
So gradient of radius is (6-0)/(2-10)
= -3/4.
?
Since 4/3 x -3/4 = -1, radius is perpendicular to tangent.
P
Intersections!
How could you tell if a line and a circle intersect:
0 times
once
twice
(x+2)2 + y2 = 33
Equate the expressions then look at the discriminant:
b2 – 4ac
? <0
b2 – 4ac
? >0
b2 – 4ac
? =0
This allows us to prove that a
line is a tangent to the circle.
Completing the square
[Edexcel] The circle C, with centre at the point A, has equation
x2 + y2 – 10x + 9 = 0. Find
(a)
the coordinates of A,
(2)
(b)
the radius of C,
(2)
x2 – 10x + y2 + 9 = 0
(x – 5)2 – 25 + y2 + 9 = 0
(x – 5)2 + y2 = 16
?
So centre is (5,0), radius is 4.
Equation of a circle
Equation
Centre
x2 + y2 + 4y – 5 = 0
(0,?-2)
x2 + y2 – 6x + 4y – 3 = 0
(3,?-2)
x2 + y2 + 12x + 2y + 12 = 0
(-6,?-1)
x2 + y2 – 4x – 6y = 3
(2,?3)
x2 + y 2 + x + y = 1
(-0.5,?-0.5)
Radius
?3
?4
?5
?4
?
3/2
(Note: this appears in exams, but not in your textbook!)

similar documents