Harder Exercise on Equations of Circles
1.
y
Figure 3
C
Q
O
x
L
R
P
In Figure 3, the line L: y = 2x + c cuts the circle C: (x + 1)2 + (y − 1)2 = 16 at the two points
and Q(x2, y2) where c < 0.
(a) Show that
( x1 − x2 ) 2 = −
4 2
(c − 6c − 71) .
25
P(x1, y1)
(4 marks)
(b) R is a point such that O is the mid-point of QR.
(i) By considering the relation between the coordinates of Q and R, show that
PR =
2
5(c 2 + 1) .
5
(ii) If PR = 2, find the area of PQR.
(8 marks)
2.
3.
4.
1.(a)
(1)
 y = 2x + c

2
2
( 2)
( x + 1) + ( y − 1) = 16
Subs. (1) into (2)
(x + 1)2 + (2x + c − 1)2 = 16
x2 + 2x + 1 + 4x2 + 4(c − 1)x + (c − 1)2 − 16 = 0
5x2 + 2(2c − 1)x + (c2 − 2c − 14) = 0
2(2c − 1)
x1 + x 2 = −
5
2
c − 2c − 14
x1 x 2 =
5
2
( x1 − x 2 )
1A
1A
= x12 − 2 x1 x 2 + x 22
= x12 + 2 x1 x 2 + x 22 − 4 x1 x 2
1M
= ( x1 + x 2 ) 2 − 4 x1 x 2
 c 2 − 2c − 14 
 2( 2c − 1) 


= −
−
4

5
5




2
=
(
)
4
4c 2 − 4c + 1 − 5c 2 + 10c + 70
25
4
= − (c 2 − 6c − 71)
25
Since OR = OQ, Q and R are symmetric about the origin.
R = (−x2, −y2)
=
1.(b)(i)
 c 2 − 2c − 14 
4( 4c 2 − 4c + 1)

− 4
25
5


PR = ( x1 + x2 ) + ( y1 + y2 )
2
1A
1A
2
= ( x1 + x2 ) + ( 2 x1 + c + 2 x2 + c )
2
1M
2
= ( x1 + x2 ) 2 + [ 2( x1 + x2 ) + 2c ]2
2

 2( 2c − 1)    2( 2c − 1) 
= −
 +  2 −
 + 2c 
5
5

  


=
4( 4c 2 − 4c + 1)
 − 4c + 2 
+ 4
+ c
25
 5

2
4c 2 − 4c + 1 + ( c 2 + 4c + 4)
5
2
=
5(c 2 + 1)
5
=
2
1A
2
1A
1.(b)(ii)
PR = 2
2
5c 2 + 5 = 2
5
5c 2 + 5 = 25
c2 = 4
c = −2
( c  0)
Area of Δ𝑃𝑄𝑅
= Area of Δ𝑄𝑅𝑇 − Area of Δ𝑃𝑄𝑆 − Area of 𝑃𝑅𝑇𝑆
1
1
1
= (2𝑥2 )(2𝑦2 ) − (𝑥2 − 𝑥1 )(𝑦2 − 𝑦1 ) − [(𝑥2 − 𝑥1 ) + (2𝑥2 )](𝑦1 + 𝑦2 )
2
2
2
1
= (4𝑥2 𝑦2 − 𝑥2 𝑦2 + 𝑥2 𝑦1 + 𝑥1 𝑦2 − 𝑥1 𝑦1 − (3𝑥2 − 𝑥1 )(𝑦1 + 𝑦2 ))
2
1
= (3𝑥2 𝑦2 + 𝑥2 𝑦1 + 𝑥1 𝑦2 − 𝑥1 𝑦1 − 3𝑥2 𝑦1 − 3𝑥2 𝑦2 + 𝑥1 𝑦1 + 𝑥1 𝑦2 )
2
1
= (−2𝑥2 𝑦1 + 2𝑥1 𝑦2 )
2
= 𝑥1 𝑦2 − 𝑥2 𝑦1
= 𝑥1 (𝑥2 − 2) − 𝑥2 (𝑥1 − 2)
= 2(𝑥2 − 𝑥1 )
= 2√−
4
((−2)2 − 6(−2) − 71)
25
4√55
5
Centre of 𝐶2 = (11, −8)
Radius of 𝐶2 = 7
Distance between the centres
= √(11 − 5)2 + (−8 − 0)2
= 10
Radius of 𝐶1
= 10 − 7
=3
∴ The equation of 𝐶1 is (𝑥 − 5)2 + 𝑦 2 = 9.
Let the equation of the tangent by 𝑦 = 𝑚𝑥.
(𝑥 − 5)2 + (𝑚𝑥)2 = 9
(𝑚2 + 1)𝑥 2 − 10𝑥 + 16 = 0
Δ=0
2
2
(−10) − 4(𝑚 + 1)(16) = 0
25 − 16𝑚2 − 16 = 0
16𝑚2 = 9
3
𝑚=±
4
=
2.(a)
2.(b)
3
∴ The equations of the tangents are 𝑦 = ± 𝑥.
4
1A
1M
1A
1A
2.(c)
3.(a)(i)
3.(a)(ii)
3.(a)(iii)
3
The equation of the tangent is 𝑦 = − 𝑥. (2)
4
Subs. (2) into (𝑥 − 11)2 + (𝑦 + 8)2 = 49
2
3
(𝑥 − 11)2 + (− 𝑥 + 8) = 49
4
9 2
𝑥 2 − 22𝑥 + 121 +
𝑥 − 12𝑥 + 64 − 49 = 0
16
25 2
𝑥 − 34𝑥 + 136 = 0
(∗)
16
Let 𝐴 = (𝑥1 , 𝑦1 ) and 𝐵 = (𝑥2 , 𝑦2 ).
𝑥1 and 𝑥2 are the roots of (*)
34 544
𝑥1 + 𝑥2 =
=
25
25
16
𝑥1 + 𝑥2
𝑥-coordinate of the mid-point =
2
1 544
)
= (
2 25
272
=
25
3 272
)
𝑦-coordinate of the mid-point = − (
4 25
204
=−
25
272 204
).
∴ The coordinates of the mid-point of 𝐴𝐵 = (
,−
25
25
∵ ∠𝐶𝐴𝐸 = 90°
∴ ∠𝐶𝐴𝐸 + ∠𝐹𝐸𝐴 = 180°
Hence 𝐴𝐵//𝐸𝐹
(int. ∠s supp.)
(vert. opp. ∠s)
∵ ∠𝐹𝐷𝐸 = ∠𝐶𝐷𝐵
(base ∠s, isos. Δ)
∠𝐶𝐷𝐵 = ∠𝐶𝐵𝐷
(alt. ∠s, 𝐴𝐵//𝐸𝐹)
∠𝐶𝐵𝐷 = ∠𝐹𝐸𝐷
∴ ∠𝐹𝐷𝐸 = ∠𝐹𝐸𝐷
(sides opp. equal ∠s)
Hence 𝐹𝐷 = 𝐹𝐸
Let 𝐶 be the circle passing through 𝐷 and touching 𝐴𝐸 at 𝐸.
∵ 𝐶 touches 𝐴𝐸 at 𝐸 and 𝐸𝐹 ⊥ 𝐴𝐸
∴ the centre of 𝐶 lies on the line 𝐸𝐹.
∵ 𝐸𝐷 is a chord of 𝐶 and 𝐹𝐷 = 𝐹𝐸
∴ the centre of 𝐶 lies on the perpendicular of 𝐷𝐸 through 𝐹.
𝐹 is the intersecdtion of the lines which is the cnetre of 𝐶.
3.(b)
Mid-point of 𝐷𝐸 = (−3, 3)
∵ 𝐸𝐷 is horizontal
∴ 𝑥-coordinate of 𝐹 = −3
Slope of 𝐴𝐸 = −2
Equation of 𝐸𝐹:
1
𝑦 − 3 = (𝑥 + 4)
2
𝑥 − 2𝑦 + 10 = 0
Subs. 𝑥 = −3 into 𝐸𝐹
−3 − 2𝑦 + 10 = 0
7
𝑦=
2
4.
7
∴ 𝐹 = (−3, )
2