n個,求(m,n) =
0 (-1,3) 8 = √8
(X + 1)² + ( Y-³)^² = 8
0 (-1₁3) X=√√8
(4,3)
形成直角三角形,求k=
12k + ²k +6²
JK² +9
=
5x-1-20
FX-1-3-0
5/1
Y+3=mx
mx -y + (-1m - 2) =
Y+2= M(x-1)
12.在圓C:x²+y^-6x+by-15=0上取三點 P(0,-3)、A(7,-2)、B,若∠APB=90°,求B點
座標
mx
0 ( ³ , - \ / ) , Y = √5 + 9 + 1 f
15
74
mx-y
mx-Y
MX-Y-3=0
(@2,0) = √8
13. 已知直線L:kx+3y+2k+6=0與圓C:x²+y^-4x-4=0交於AB兩點,且兩點恰與圓心
2
=> 14k +6| = √√8K² + 72
=> 16k² + 48 k +36 = 8k² + 12
2
8k² + 48k-36=0
2k ²+12k - Y=0
k+ 6k-6=0
k
k
7+3=mx
-1
6
36 - 24k =0
k=
36
fla
TX-Y+(-5-2)
5-74-49:0
4
(20)
=0
100