A.Maths高手請來 [復制鏈接]

我投降....係忘記晒,...真係 "還返晒比老師" 啦

我計到h= 3/2
r= square root 5/2
有冇錯到

correct

Then find the equation of the circle

都係錯...唉..冇哂心機

引用第32樓kylau於2007-05-15 01:08發表的“”:
correct
Then find the equation of the circle

點解計唔到既

你睇下係唔係咁

(y-3/2)^2 + x^2= 5/4

引用第33樓樂仔於2007-05-15 01:08發表的“”:
都係錯...唉..冇哂心機

(書個答案未必岩的)

我最後計到x^2+y^2-3y+1=0...但係ans唔係咁

I've made a typo mistake

So use slope between (h,0) and (1,1) = -1/2

Should be (0,h) instead of (h,0).

U try once more, sorry

引用第37樓kylau於2007-05-15 01:11發表的“”:
I've made a typo mistake [表情]
So use slope between (h,0) and (1,1) = -1/2
Should be (0,h) instead of (h,0).
.......

你無錯丫
個center係(h,0)喎

死喇...愈整愈亂

Check if h=3, r = root 5

eq is x^2 + y^2 - 6x + 4 =0
--------------------
slope between (h,0) and (1,1) = -1/2

1 / (1-h) = -1/2

h=3

哦..明喇

h應該係=3..right?

yes, Check if eq is x^2 + y^2 - 6x + 4 =0

引用第39樓kylau於2007-05-15 01:13發表的“”:
Check if h=3, r = root 5
eq is x^2 + y^2 - 6x + 4 =0
--------------------
slope between (h,0) and (1,1) = -1/2
.......

搞掂

thx!!

呢條其實係第20條...之後既第21條至到26條...通通都唔識

21) Find the equation of the circle which passes through the point (1, 2) and touches the line x-3y-5=0 at (-1, -2)