maths~~~~~

(a) (i) 代 x = 0, 則:

y2 + 8y + 16 = 0

y = -4

所以: Q 為 (0, -4)

代 y = 0, 則:

x2 - 8x + 16 = 0

x = 4

所以: P 為 (4, 0)

(ii) 圓半徑 = 4

EO = 4√2

OS = 4√2 + 4

OR = 4√2 - 4

即 S 為: (4√2 + 4, -4√2 - 4)

R 為: (4√2 - 4, -4√2 + 4)

(b) 設 L 與 C2 的另一交點為 Q, 則以相似圖形計:

OS/OR = OR/OQ

OQ = (4√2 - 4)2/(4√2 + 4)

= 4(5√2 - 7)

RQ = 4√2 - 4 - 4(5√2 - 7) = 8(3 - 2√2), 即 C2 半徑 = 4(3 - 2√2)

Q 在 [8(3 - 2√2), -8(3 - 2√2)]

C2 圓心在 (20 - 12√2, 12√2 - 20)

C2 方程式:

[x - (20 - 12√2)]2 + [y - (12√2 - 20)]2 = [4(3 - 2√2)]2

x2 + y2 - (40 - 24√2)x - (24√2 - 40)y + 32(43 - 30√2) = 16(17 - 12√2)

x2 + y2 + 8(3√2 - 5)x + 8(5 - 3√2)y + 48(23 - 16√2) = 0

a) i)
C1: x^2+y^2-8x+8y+16=0

Put x = 0,
y = -4
Put y = 0,
x = 4

Therefore, P =(4,0) and Q =(0,-4).

a) ii)
E =(4,-4)
OE = sqrt.[(4)^2+(-4)^2] = 4sqrt.2
Therefore, OR = OE - ER = 4(sqrt.2 - 1)
...............OS = OE +ES = 4(sqrt.2 +1)

b) Let the radius of C2 be r,
By similarly,
OR = r(sqrt.2 +1) = 4(sqrt.2 - 1)
.........r = 4(sqrt.2 - 1)/(sqrt.2 +1)
...........= 4(sqrt.2 - 1)^2
...........= 8 - 8sqrt.2 +4
...........= 12 - 8sqrt.2
centre of C2 = (r,-r)
Therefore,
equation of C2: (x - 12+8sqrt.2)^2 + (y +12 - 8sqrt.2)^2 = 16(3 - 2sqrt.2)^2
.......................(x - 12+8sqrt.2)^2 + (y +12 - 8sqrt.2)^2 = 144 - 192sqrt.2 +128
.......................(x - 12+8sqrt.2)^2 + (y +12 - 8sqrt.2)^2 = 272 - 192sqrt.2

做完喇...
希望幫到你!(錯左同我講聲喎..因為D數好易計錯.)