### The circle

```Coordinate Geometry
of the line questions
The circle C has the equation
x 2 + y 2 = 49
(i) Write down the centre and radius of C?
Centre = (0,0)
(ii) Verify that the point (5, -5) lies outside C
(5)2 + (-5)2 = 25 + 25 = 50 > 49 so it is outside
(b) The Line y = 10 – 2x intersects with the circle x 2 + y 2 = 40
Find the points of intersection?
• x 2 + (10 – 2x) 2 = 40
• x 2 + 100 – 40x + 4x 2 = 40
• 5x 2 – 40x + 60 = 0
• x – 8x + 12 = 0
•(x – 6)(x – 2) = 0
•Solutions: x = 6 or 2
•Corresponding y values?
• y = 10 – 2(6) = -2
• y = 10 – 2(2) = 6
•Answers: (6 , -2) and (2 , 6)
2002 LC Paper 2 Q. 3
(0, 10)
10
(-10, 0)
10
(10, 0)
10
10
x2 + y2 = 100
Centre (0, 0) |Radius| = 10
Points on circle:
(10, 0), (0, 10) (-10, 0)
Equation of Circle C: (x - 2)2 + (y + 1)2 = 8
Centre (2, -1) |Radius| =  8
C cuts y-axis when x = 0
(y + 3)(y – 1) = 0
(0 - 2)2 + (y + 1)2 = 8
y + 3 = 0 or y – 1 = 0
4 + y2 + 2y + 1 = 8
y2
y2
+ 2y + 5 - 8 = 0
+ 2y - 3 = 0
y = - 3 or y = 1
C cuts y-axis at
a(0, -3) and b(0, 1)
(x – 2)2 + (y + 1)2 = 8
T
Centre (2, -1) |Radius| =  8
RT
y2 – y1
Slope of R =
x2 – x1
x1 = 2 x2 = 4
y1 = -1 y2 = 1
(1) – (-1)
(2, -1)
R
(4, 1)
=
(4) – (2)
2
=
2
=1
Slope of T = - 1
Point on T (4, 1)
Slope of T = -1
Equation of T: y – y1 = m(x – x1)
x1 = 4 m = -1
y1 = 1
y – (1) = -1(x – (4))
y–1=-x+4
x + y – 5 = 0 Equation of T
8
b(3, 7)
7
6
5
4
3
2
a(-5, 1)
1
0
-6
-4
-2
0
2
4
6
8
-1
-2
From the diagram the right-angle appears to be at b
We have to show that ab  bc
10
c(9, -1)
y2 – y1
Slope of ab =
=
=
x2 – x1
(7) – (1)
(3) – (-5)
6
8
3
=
4
x1 = - 5
y1 = 1
x2 = 3
y2 = 7
Slope of bc =
=
=
y2 – y1
x2 – x1
(-1) – (7)
(9) – (3)
-8
6
-4
=
3
x1 = 3
y1 = 7
x2 = 9
y2 = -1
Slope of ab =
3
4
Product of slopes =
Slope of bc =
3
x
4
4
3
 ab  bc
 abc is right-angled at b
= -1
-4
3
b(3, 7)
Centre of Circle, d,
is the mid-point of [ac]
a(-5, 1)
d
c(9, -1)
abc = 900 is an angle in a semi-circle
of which [ac] is the diameter
a(-5, 1)
d(2, 0)
c(9, -1)
 x1  x2 y1  y2 
Mid-point of [ac] = 2 , 2  


x1 = - 5 x2 = 9
y1 = 1 y2 = -1
 (5)  (9) (1)  (1) 
,


2
2


  5  9 1 1 
,


2
2


d 2, 0
= Centre of Circle
|Radius| = Distance from d(2, 0) to a(-5, 1)
x2  x1 2  y2  y1 2
| ac| 
x1 = -5 x2 = 2
y1 = 1 y2 = 0
2  (-5) 2  0  12


7 2  12
 49  1
Centre = (2, 0) |Radius| 50
Equation of Circle:
(x - 2)2 + (y)2 = 50
2001 Paper 2 Question 3
S: (x – 3)2 + (y – 4)2 = 25
Centre: (3, 4) |Radius| = 5
(k, 0) is on the circle S
 k(k – 6) = 0
 (k – 3)2 + (0 – 4)2 = 25
 k = 0 or k - 6 = 0
 k2 – 6k + 9 + 16 = 25
 k = 0 or k = 6
 k2 – 6k + 25 – 25 = 0
 k2 – 6k = 0
The line is a tangent if there is only one point of intersection
x – 3y = 10
x2 + y2 = 10 Substitute 3y + 10 for x
x = 3y + 10
(3y + 10)2 + y2 = 10
9y2 + 60y + 100 + y2 = 10
10y2 + 60y + 90 = 0
Divide by 10
y2 + 6y + 9 = 0
The line is a tangent if there is only one point of intersection
x – 3y = 10
y2 + 6y + 9 = 0
x = 3y + 10
(y + 3)(y + 3) = 0
When y = -3
x = 3(-3) + 10
x = - 9 + 10 = 1
y + 3 = 0 or y + 3 = 0
y=-3
Only one point of intersection (1, -3)
 The line is a tangent
o(0, 0) a(1, -5)
| oa | 
x2  x1 2  y2  y1 2

1  02  - 5  02

12  - 52
Centre = (0, 0) |Radius|  26
Equation of C: x2 + y2 = 26
 1  25
 26
x1 = 0 x2 = 1
y1 = 0 y2 = -5
(p, p) First Component = Second Component
is on the line y = x
6
5
PZ
(4,4)
4
(3,3)
3
(2,2)
2
1
-6
-5
-4
-3
0 (0,0)
-1 (-1,-1)
0
1
-1
-2
(-2,-2) -2
(-3,-3)
(-4,-4)
-3
-4
-5
-6
(1,1)
2
3
4
5
6
Values of p are
-3, -2, -1, 0, 1, 2, 3
2000 Paper 2 Question 3
x2 + y2 = 16 is the circle C
Centre (0, 0) and |Radius| = 4
o(0, 0) a(3, 1)
x2  x1 2  y2  y1 2
| oa | 

3  02  1  02

32  12
 9 1
 10 = 3.16 < |Radius|
a(3,1) is inside the circle
x1 = 0 x 2 = 3
y1 = 0 y2 = 1
x2 + y2 = 29
Centre (0, 0) |Radius| = 29
T
(0, 0)
R
(2, 5)
RT
y2 – y1
Slope of R =
x2 – x1
=
(5) – (0)
(2) – (0)
5
=
2
5
= 2
2
Slope of T= - 5
x1 = 0 x2 = 2
y1 = 0 y2 = 5
Point on T (2, 5)
2
Slope of T= - 5
Equation of T: y – y1 = m(x – x1)
x1 = 2
y1 = 5
m = -2
5
y – (5) = -2 (x – (2))
5
y - 5 = -2 (x - 2)
5
Multiply by LCD = 5
5y - 25 = (5)(-2) (x - 2) = - 2x + 4
5
2x + 5y - 29 = 0 Equation of T
a(-2, -3)
b(-4, 3)
Centre of Circle
is the mid-point
of the diameter ab
a(-2, -3)
c(-3, 0)
b(- 4, 3)
 x1  x2 y1  y2 
Mid-point of [ab] = 2 , 2  


x1 = - 2 x2 = - 4
y1 = - 3 y2 = 3
 (2)  (4) (3)  (3) 
,


2
2


  2  4  3  3    6 , 0   c - 3, 0 
,


 
2
2
2
2


 
= Centre of Circle
|Radius| = Distance from c(-3, 0) to a(-2, -3)
| ac| 
x2  x1 2  y2  y1 2

- 3  - 22  0  - 32

- 3  22  0  32

- 12  32
 19
x1 = -2 x2 = -3
y1 = -3 y2 = 0
Equation of Circle:
(x + 3)2 + (y)2 = 10
Equation of Circle: (x + 3)2 + y2 = 10
Circle cuts y-axis when x = 0
(0 +
3)2
+
y2
= 10
9 + y2 = 10
y2 = 10 – 9 = 1
y2
–1=0
(y + 1)(y – 1) = 0
y + 1 = 0 or y – 1 = 0
y = - 1 or y = 1
Circle cuts y-axis at
a(0, -1) and b(0, 1)
These points on the y-axis
are 2 units apart
|ab| = 2
From an Accurate diagram
3
2
6
c
-6
b(0, 1)
1
-5
-4
20
-3
-2
-1
0
-1
d
Area of•abcd = (6)(2) = 12
-2
-3
1
a(0, -1)
2
3
1999 Paper 2 Question 3
o(0, 0) a(8, 6)
x2  x1 2  y2  y1 2
| oa | 

8  0 2  6  0 2

82  62
 64  36
 100  10 = |Radius|
x1 = 0 x2 = 8
y1 = 0 y2 = 6
Centre = (0, 0) |Radius| = 10
Equation of C: x2 + y2 = 100
a(-1, -1)
b(3, -3)
Centre of S
is the mid-point
of the diameter ab
a(-1, -1)
c(1, -2)
b(3, -3)
x1 = - 1 x2 = 3
y1 = - 1 y2 = - 3
Mid-point of [ab]
 x1  x2 y1  y2 
= 2 , 2  
  1  3  1  (3) 
,


2
 2

  1  3  1  3  c 1,2
,


2
2


= Centre of S
|Radius| = Distance from c(1, -2) to a(-1, -1)
| ac| 
x2  x1 2  y2  y1 2

- 1  12  - 2  (1)2

- 22  - 2  12
Centre = (1, -2)|Radius|  5
Equation of S:
(x – 1)2 + (y + 2)2 = 5
 4 1
 5
x1 = 1 x2 = -1
y1 = -2 y2 = -1
x2 + y2 = 13
Centre (0, 0) |Radius| = 13
T
RT
y2 – y1
Slope of R =
x2 – x1
(-3) – (0)
(0, 0)
R
(-2, -3)
=
(-2) – (0)
-3
=
-2
3
= 2
2
Slope of T= - 3
x1 = 0 x2 = -2
y1 = 0 y2 = -3
2
Point on T (-2, -3) Slope of T= - 3
Equation of T: y – y1 = m(x – x1)
x1 = -2
y1 = -3
m = -2
3
y – (-3) = -2 (x – (-2))
3
y + 3 = -2 (x +2)
3
Multiply by LCD = 3
3y + 9 = (3)(-2) (x +2) = - 2x - 4
3
2x + 3y + 13 = 0 Equation of T
T
(0, 0)
2
Slope= - 3
R
The point needed is
(-2, -3) the other end of the
diameter as shown
2
Slope= - 3
The other end of
the diameter is
(2, 3)
(0, 0)
T
(-2, -3)
The translation that maps (-2,-3)  (0, 0)
will map (0, 0)  ( 2,3 )
Point (2, 3)
2
Slope of T= - 3
Equation of || tangent: y – y1 = m(x – x1) x1 = 2
y1 = 3
m = -2
3
y – (3) = -2 (x – (2))
3
y - 3 = -2 (x - 2)
3
Multiply by LCD = 3
3y - 9 = (3)(-2) (x - 2) = - 2x + 4
3
2x + 3y - 13 = 0 Equation of other tangent
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