Question

求这两题详细答案

Answer 1

C1：(x-4)^2+(y-3)^2=r^2
ρsin(θ-π/4)=m
ρ(sinθcosπ/4-cosθsinπ/4)=m
√2/2(ρsinθ-ρcosθ)=m
y-x=√2m
C2：y=x+√2m
(1) r=1
(x-4)^2+(y-3)^2=1
(x-4)^2+(x+√2m-3)^2=1

2x^2+(-8+2√2m-6)x+(√2m-3)^2-1=0
2x^2+(2√2m-14)x+2m^2-6√2m+8=0
x^2+(√2m-7)x+m^2-3√2m+4=0
∵只有一个交点
∴Δ=0
(√2m-7)^2-4(m^2-3√2m+4)=0
2m^2-14√2m+49-4m^2+12√2m-16=0
-2m^2-2√2m+33=0
m^2+√2m=33/2
(m+√2/2)^2=33/2+1/2
m+√2/2=±√17
m=-√2/2±√17
(2) m=3-√2/2

C2：y=x+√2(3-√2/2)

y=x+3√2-1
(x-4)^2+(y-3)^2=r^2
x^2-8x+16+(x+3√2-1-3)^2=r^2
x^2-8x+16+x^2+2(3√2-4)x+(3√2-4)^2-r^2=0
2x^2+(6√2-16)x+16+18-24√2+16-r^2=0
x^2+(3√2-8)x+25-12√2-1/2r^2=0
A(x1,y1)、B(x2,y2)

x1+x2=8-3√2
x1x2=25-12√2-1/2r^2
(x1-x2)^2=(x1+x2)^2-4x1x2
=(8-3√2)^2-4(25-12√2-1/2r^2)
=64-48√2+18-100+48√2+2r^2
=-18+2r^2
|AB|=√[(1+k^2)(x1-x2)^2]
=√[(1+1^2)(-18+2r^2)]
=√(-36+4r^2)
|AB|=8
√(-36+4r^2)=8
-36+4r^2=64
r^2=25
r=±5(负号舍去)
r=5
注：^2——表示平方。