YLL討論網

#ed_op#DIV#ed_cl##ed_op#IMG alt="image file name: 2k3d83f9c1fb.png" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2k3d83f9c1fb.png" border=0#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(x-Cx)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# + (y-Cy)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# = BC#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#BR#ed_cl#(x-0)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# + (y-5)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# = 4.5#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# + y#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# - 10y + 19/4 = 0   .... (1)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(x-Ax)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# + (y-Ay)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# = AB#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#BR#ed_cl#(x-0)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# + (y-0)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# = 5.5#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#BR#ed_cl#x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# + y#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# = 121/4   ............. (2)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# = 121/4 - y#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#上式代入 (1) 式#ed_op#BR#ed_cl#121/4 - y#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# + y#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# - 10y + 19/4 = 0#ed_op#BR#ed_cl#140/4 - 10y = 0#ed_op#BR#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#y = 14/2 = 7/2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#上式代入 (2) 式#ed_op#BR#ed_cl#x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# + (7/2)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# = 121/4#ed_op#BR#ed_cl#x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# + 49/4 = 121/4#ed_op#BR#ed_cl#x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# = (121-49)/4 #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# = 72/4 = 18#ed_op#BR#ed_cl#x = 3·√2,  -3·√2 (此題只要正, 故負不合).#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

要求出座標(3根號2,3.5)沒問題
要公式!?

有三個圓(a,b,c)直俓分別為4、5、6
其中有(c)圓(直徑6)在座標(0,0)
(a)圓(直徑4)在座標(0,6/2 + 4/2)=(0,5)
(b)圓(直徑5)必須緊鄰(a)圓及(c)圓且座標要為正
依這個條件是否有公式能求(b)圓的座標呢??