A.Maths高手請來 [復制鏈接]

引用第13樓樂仔於2007-05-15 00:50發表的“”:
係唔係即係咁
(1-h)^2+(-1)^2= d

ya..

個center同條line既distance

有條公式搵架...應該係代入去..睇下書啦..應該有寫

引用第14樓垃圾桶於2007-05-15 00:50發表的“”:
ya

都係要...in terms 一d野

引用第15樓ｊ摩連奴．狂妄於2007-05-15 00:52發表的“”:
ya..
個center同條line既distance
.......

我有背條公式...依家試緊代..唔該你先

做唔到...

咁淺我都唔記得

都唔知我點考 PURE MATH C返黎

Centre of circle at (h,0)

So eq of circle is (x-h)^2 + y^2 = r^2 .............(*)

Intercept with 2x-y-1=0 at (1,1)

So sub (1,1) into (*) get one equation

For line 2x-y-1=0, slope = 2

So use slope between (h,0) and (1,1) = -1/2 to have the second equation.

Solve for h first, then r, then eq of circle is done.

引用第19樓ballballfull於2007-05-15 00:58發表的“”:
咁淺我都唔記得 [表情]
都唔知我點考 PURE MATH C返黎 [表情]

超勁...我果時a maths緊緊合格.....

引用第20樓kylau於2007-05-15 00:58發表的“”:
Centre of circle at (h,0)
So eq of circle is (x-h)^2 + y^2 = r^2 .............(*)
Intercept with 2x-y-1=0 at (1,1)
So sub (1,1) into (*) get one equation
For line 2x-y-1=0, slope = 2
So use slope between (0,h) and (1,1) = -1/2 to have the second equation.
Solve for h first, then r, then eq of circle is done.
.......

呢度開始唔明

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

呢2點之間個slope係-1/2 x埋90 degree

2 point form就搵到h

之後

哦...兩個slope乘埋=-1 ..明白

引用第22樓樂仔於2007-05-15 01:00發表的“”:
呢度開始唔明 [表情]

two line, 90 degree

引用第22樓樂仔於2007-05-15 01:00發表的“”:
呢度開始唔明 [表情]

For line 2x-y-1=0, slope = 2

So use slope between (0,h) and (1,1) = -1/2 to have the second equation.

兩個slope成90度....乘埋=-1

tangent is prependicular to radius,

So ....

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

引用第27樓kylau於2007-05-15 01:02發表的“”:
tangent is prependicular to radius,
So ....
slope between (0,h) and (1,1) = -1/2

嗯...我依家搵到個h出黎

之後就搵r?

引用第28樓樂仔於2007-05-15 01:04發表的“”:
嗯...我依家搵到個h出黎
之後就搵r?

當然啦

有center了，center同1,1.....個距離仲唔係radius