F.5 Math's Q

呢啲加奶

2013-04-19 4:38 am

唔洗全部都做，如果你識做，才全部都做。

New Senior Secondary Mathematics in Action Complusory Part
Chapter5 Equations of Circles:

1. The circle of radius r touches both the x-axis and the y-axis, and it passes
through the point A(-1,8). If C is the centre of the circle, find

(a) the coordinates of C in terms of r, Ans: (-r, r)

(b) the possible equations of the circle. Ans: (x+5)^2 + (y+5/2)^2 = 25/4

2.A circle with centre C cuts the x-axis at two points A(2,0) and B(8,0) and
touches the y-axis. The point P(1, -2) lies inside the cicle.

(a) Find the x-coordinate of C and the radius of the circle.
Ans: x-coordinate of C =5, radius=5

(b) Hence, find the equation of the circle. Ans: (x-5)^2 + (y+4)^2 =25

3. THe circle C1 : x^2+y^2-8x+4y+15=0 cuts the c-axis at two point A(3,0) and
B(5,0). S is the centre of the circle C1. Find

(a) the equation of the circle C2 passing through A, B and S,
Ans: 2x^2+2y^2 -16x+3y+30=0

(b) the ration of the area of C1 to that of C2. Ans: 16:5

4. The triangle formed by the three points A(2,2), B(6,4) and C(10, -4).
Triangle ABC.

Find the coordinates of the point D lying on the circle (x^2+y^2-12x+2y+12=0)
such that ABCD is a rectangle.

Ans:(6,-6)

5. The circle C: x^2+y^2+4x+5y+4=0 cuts the y-axis at two points P(0,-1)and
Q(0,-4) and touches the x-axis at the point R(-2,0). The line passing through P
and perpendicular to QR intersects QR at the point S.

Find the coordinates of S. Ans: (-6/5, -8/5)

回答 (1)

fooks

2013-04-19 7:24 am

✔ 最佳答案

1.
(a)
The circle touches the x-axis and the y-axis.
|the x coordinate of C| = |the y coordinate of C|= r

The circle passes through A(-1, 8).
C is in the 2nd quadrant.
Hence, C = (-r, r)

(b)
r = CA
r = √[(-r + 1)² + (r - 8)²]
r² - 18r + 65 = 0
(r - 5)(r - 13) = 0
r = 5 or r = 13

When r = 5, the equation of the circle :
(x + 5)² + (y - 5)² = 25

When r = 13, the equation of the circle :
(x + 13)² + (y - 13)² = 169

(The given answer is incorrect.)

2.
(a)
x-coordinate of C
= (2 + 8)/2
= 5

Since the circle touches the y-axis, radius = 5

(b)
Let (5, b) be C.
CA = 5
√[(5 - 2)² + (b - 0)²] = 5
b² = 16
b = 4 (rejected) or b = -4

The equation of the circle :
(x - 5)² + (y + 4)² = 25

3.
(a)
S = (-[-8/2], -4/2) = (4, -2)

Let C2 : x² + y² + Dx + Ey + F = 0

Put (3, 0), (5, 0) and (4 , -2) into C2 :
9 + 3D + F = 0 ...... [1]
25 + 5D + F = 0 ...... [2]
16 + 4 + 4D - 2E + F = 0 ...... [3]

[2] - [1] :
16 +2D = 0
D = -8

Put D = -8 into [1] :
9 + 3(-8) + F = 0
F = 15

Put D = -8 and F = 15 into [3] :
20 + 4(-8) - 2E + 15 = 0
E = 3/2

C2 : x² + y² - 8x + (3/2)y + 15 = 0
C2 : 2x² + 2y² - 16x + 3y + 30 = 0

(b)
radius of C1 = √[(-8/2)² + (4/2)² - 15] = √5
radius of C2 = √[(-16/4)² + (3/4)² - (30/2)] = 5/4

Area of C1 : Area of C2
= π(√5)² : π(5/4)²
= 5 : 25/16
= 16 : 5

4.
Let (h, k) be D.

The diagonals of a rectangle bisect each other.
(h + 6)/2 = (2 + 10)/2 ...... [1]
(k + 4)/2 = (2 - 4)/2 ...... [2]

From [1] :
h + 6 = 12
h = 6

From [2] :
k + 4 = -2
k = -6

D = (6, -6)

5.
QR:
(y + 4)/(x - 0) = (0 + 4)/(-2 - 0)
2x + y + 4 = 0 ...... [1]

Slope of QR = -2
Slope of the line passing through P and perpendicular to QR :
(y + 1)/(x - 0) = -1/(-2)
x - 2y - 2 = 0 ...... [2]

[1]*2 + [2] :
5x + 6 = 0
x = -6/5

[1] - [2]*2 :
5y + 8 = 0
y = -8/5

Hence, S = (-6/5, -8/5)

參考: fooks

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