關於F5 解幾"圓"的問題!

Q1. incomplete question

Q2. x+y+c=0 ....(1)
x^2+y^2=1....(2)

from (1), y=-x-c ....(3)
put (3) into (2) , x^2+(-x-c)^2=1
i.e. 2x^2+2cx+(c^2-1)=0 ....(4)

since two distinct point , delta of (4) >0
(2c)^2-4(2)(c^2-1)>0
c^2-2<0
-√2<c<√2

Q3. Let the required circle be:
x^2+y^2-4+K(x^2+y^2+6x)=0 (where K is constant and not equal -1) ...(1)
since (-2,-2) on (1)
put (-2,-2) on (1)
we get 4+4-4+K(4+4-12)=0
K=1
the the required circle be x^2+y^2+3x-2=0