A.Maths高手請來

引用第4樓ｊ摩連奴．狂妄於2007-05-15 00:42發表的“”:
x^2+y^2+Dx+Ey+F
用個center(1, 1)搵到D,E...
跟住個radius=個center同條line既distance...搵埋F..

引用第5樓樂仔於2007-05-15 00:42發表的“”:
連個radius點搵都唔知 [表情]

引用第6樓樂仔於2007-05-15 00:43發表的“”:
center唔係(1, 1)丫 [表情]

引用第8樓樂仔於2007-05-15 00:46發表的“”:
我連題目都未解得掂
Find the equation of the cirlce {with the center on the x-axis} {and touching the line 2x-y-1=0 at (1, 1)}
or
.......

引用第9樓ｊ摩連奴．狂妄於2007-05-15 00:47發表的“”:
咁...
center(h,0)同(1, 1)既distance=個center同條line既distance
.......

引用第10樓ｊ摩連奴．狂妄於2007-05-15 00:48發表的“”:
上面呀 [表情]

引用第9樓ｊ摩連奴．狂妄於2007-05-15 00:47發表的“”:
咁...
center(h,0)同(1, 1)既distance=個center同條line既distance
.......

引用第11樓樂仔於2007-05-15 00:48發表的“”:
center同條line既distance應該搵唔到-0-
除非係in terms一d野...

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

引用第14樓垃圾桶於2007-05-15 00:50發表的“”:
ya

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
.......

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]

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.
.......

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

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

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

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?

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

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

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).
.......

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
.......

引用第44樓樂仔於2007-05-15 01:19發表的“”:
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)

引用第44樓樂仔於2007-05-15 01:19發表的“”:
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)

引用第46樓alvin_bun於2007-05-15 01:23發表的“”:
其實呢d來來去去都不外乎幾個方法=v=
(x-a)^2+(y-b)^2=r^2...
同
.......

引用第48樓樂仔於2007-05-15 01:26發表的“”:
第一次做功課做到咁燥...不好意思...