(x=r', y=y+u. 2=z-2u. =2+2 (=2z, 交得 x'=I, 寸=号(2+2， (2=(2g+, 推出设下垂足 +动=子+方侧+明 展开后得 25x2+4y2+22+4y2-20y-102=0. 3.股也线 〔x2+2y2+z2=1, 12+2=y 给xOy平下的论影设下垂足. 解：母线的垂向向量意(0,0,1.二得垂足别 【x=x u=. 2=2十u, x2+2/2+22=1 xP+2=, 交得 〔丈=x, （2y2+-1=0, 平于y=-1不合题是，因此设下垂足圆 y=2 2 另交：将也线垂足别的第应个垂足减去第可个垂足，二得22-y-1=0.这意母线法2等平球的 设下，及且通知两知也线，即圆所股的论影设下 4.因说明面就垂足所确向的也下意设下： (1)(x+(y+z=a: (2)(r+(+=x+2y+ (3)y2+22+2=1-x2(④(+y+22=(-y-2. 解：(1)因圆题线 {工+y=a在此也下上边的垂向圆1：(-1：1.且点（受·号·号）在此题线 ly+2=a 上及平下x-y+z-号=0法也下（红+(y+)=a2的解线圆 「(e+)(g+z)=a2 x-y+z=号 .8"    x = x 0 , y = y 0 + u, z = z 0 − 2u, y 0 = x 02 + z 02 , y 0 = 2z 0 , %=    x 0 = x, y 0 = 2 5 (2y + z), z 0 = 1 5 (2y + z), /03TU 2 5 (2y + z) = x 2 + 1 25 (2y + z) 2 , g= 25x 2 + 4y 2 + z 2 + 4yz − 20y − 10z = 0. 3. X ( x 2 + 2y 2 + z 2 = 1, x 2 + z 2 = y N xOy ;3,3TU. : TQQV (0, 0, 1). .=TU"    x = x 0 , y = y 0 , z = z 0 + u, x 02 + 2y 02 + z 02 = 1, x 02 + z 02 = y 0 , %=    x 0 = x, y 0 = y, 2y 02 + y 0 − 1 = 0, ;< y = −1 ,ra, ()3TU# y = 1 2 , Ã − √ 2 2 6 x 6 √ 2 2 ! . %: ITU"|BCTUYZ|.CTU, .= 2y 2 − y − 1 = 0. R z M;5 3, -$i:@:, t#FX,3. 4. (c34TUFPQ33: (1) (x + y)(y + z) = a 2 ; (2) (x + y)(y + z) = x + 2y + z; (3) y 2 + 2yz + z 2 = 1 − x 2 ; (4) (x + y + z) 2 = (x − y − z) 2 . : (1) (#
 ( x + y = a, y + z = a )3, ATQ# 1 : (−1) : 1. $& ³ a 2 , a 2 , a 2 ´ )
 . -;3 x − y + z − a 2 = 0 3 (x + y)(y + z) = a 2 %# ( (x + y)(y + z) = a 2 , x − y + z = a 2 . · 8 ·