Question

Xinyin

想問這題，訂正部分的1要怎麼解

y-1=m1x-47
y-mx
G. 給定圓C:x²+y²-2x+4y-4=0,已知通過P(4,7)且與圓C相切的直線有兩條1,2,
且切圓於 A,B 兩點,試求切點A,B的坐標為(4-2)or(一号,一(兩解)
設與圓C相切的L為y-7=m(x-4)
=> mx-4m-y+7=0
0 ₤ × - 2/2² -y + 7 = 0
8x-32-by + 42=0
x=4代入C
C = (x-1)²+(y+2)² = 4+1+4
⇒ 141 (1₁-2), r=3
= 9
1M-4m+2+71
=
3
Jm²+1=
19-3m 1² = 9 (m+1)
81-54m+9m² = 9m³ +9
72=54m
m =
8
6
8x-by + 10 = 0
4x-3y+5=0
14x-3y+51
5
= 3
16ty 2-8 +4y-4=0
y²+4y+4=0
(Y+ 2)² =0
y=-2

Answer 1

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阿潘

我使用你已經解出來的斜率接續解題

Xinyin

謝謝你～