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Disappeared circle (New Way, Book 2, p.101, No.4)
Yue Kwok Choy
Question
Find the equation of the circle passing through the intersections of the circles
and C2 : x2 + y2 + 2x – 4y – 4 = 0 and having area equal to 9 square units .
C1 : x2 + y2 = 4
Solution 1 (unsatisfactory)
Let the required equation of the circle be:
x2 + y2 – 4 + k(x2 + y2 + 2x – 4y – 4) = 0
(1 + k) x2 + (1 + k)y2 + 2kx – 4ky – (4 + 4k) = 0
C1 + kC2:
i.e.
x 2  y2 
2k
4k
x
y40
1 k
1 k
9k 2  8k  4
3
1  k 2


2
Area = r2 = 9
Since
…. (2)
9k 2  8k  4
 k   2k 


4


 

1  k 2
1  k  1  k 
2
Radius, r =
…. (1)

r=3.
 9k 2  8k  4  91  k 
2
 8k  4  9  18k
k
The required equation of the circle is x2 + y2 – 2x + 4y – 4 = 0
1
2
…. (3)
Solution 2
Let the required equation of the circle be:
C2 + kC1:
i.e.
(x2 + y2 + 2x – 4y – 4) + k (x2 + y2 – 4) = 0
(1 + k) x2 + (1 + k)y2 + 2x – 4y – (4 + 4k) = 0
…. (4)
2
4
x
y40
1 k
1 k
…. (5)
x 2  y2 
2
Radius, r =
Since

Area = r2 = 9
4k 2  8k  9
3
1  k 2
 k = 0 or –2 .
When k = 0,
When k = –2,

2
 1   2 

 
 4 
1  k  1  k 

4k 2  8k  9
1  k 2
r=3.
 4k 2  8k  9  91  k 
2
Using (5) , we have:
the required equation of the circle is
the required equation of the circle is
Two possible eqs:
x2 + y2 + 2x – 4y – 4 = 0 or
 5k 2  10k  0
 k k  2  0
x2 + y2 + 2x – 4y – 4 = 0 .
x2 + y2 – 2x + 4y – 4 = 0 .
x2 + y2 – 2x + 4y – 4 = 0
…. (6)
Discussion
As can be seen in
(6) , one of the solution circles is
In solution 1, the family of circles C1 + kC2 = 0

C2 : x2 + y2 + 2x – 4y – 4 = 0 .
1
C1  C 2  0
k
does not include C2
where k =  . As a result, there is only one circle for solution 1 as in
(Similarly, C2 + kC1 does not include C1 )
(3) .
Solution 3
C1 : x2 + y2 – 4 = 0
…. (7)
Their common chord is given by
C2 : x2 + y2 + 2x – 4y – 4 = 0
L : x2 + y2 – 4 = x2 + y2 + 2x – 4y – 4
…. (8)
L : x – 2y = 0
or
Let the required equation of the circle be:
C1 + kL:
(x2 + y2 – 4) + k (x – 2y) = 0
i.e.
x2 + y2 + kx – 2ky – 4 = 0
Radius, r =
k
2
  k 4 
2
2
Since
…. (9)
5k 2  16
4
Area = r2 = 9

r=3.
5k 2  16
 3  5k 2  16  36
 5k 2  20  0
 k2  4  0
4
 k = 2 or –2 .
From (9) ,
When k = 2,
the required equation of the circle is
x2 + y2 + 2x – 4y – 4 = 0 .

When k = –2,

the required equation of the circle is
Two possible equations:
Think :
x2 + y2 – 2x + 4y – 4 = 0 .
x2 + y2 + 2x – 4y – 4 = 0 or x2 + y2 – 2x + 4y – 4 = 0.
How many solution(s) if we use
(a) C2 + kL ,
(b) L + kC1 ,
(c) L + kC2
?
Solution 4
You may not use the concept of family of circles to find the solution, but it will be longer.
The working scheme is as follows:
2 
 4
,

 .
5
5

1.
Solve (7) and (8). The points of intersections are
2.
Let G(a, b) be the centre of the required circle.
The distance between G and one of the two points of contact = r = 3 .
You set up two equations involving (a, b).
Solve, you can get the centres of possible circles : (1, –2) or (–1, 2)
Using centre-radius form, the required circles are:
x2 + y2 + 2x – 4y – 4 = 0 or x2 + y2 – 2x + 4y – 4 = 0.
3.
4.
,