coordinate geometry--family of circles有困難..

2006-10-31 12:24 am

consider a circle C:x²+y²-4x+4y-24=0.A straight lineL:x-y+4=0 is tangent to C.

Show that for any value of k,the circle x²+y²-4x+4y-24+k(x-y+4)=0 touchesL.

唔該識既幫幫我=]

回答 (1)

myisland8132

2006-10-31 1:00 am

✔ 最佳答案

a circle C:x²+y²-4x+4y-24=0.A straight lineL:x-y+4=0 is tangent to C
from x-y+4=0
x=y-4...(1)
sub into C:x²+y²-4x+4y-24=0
(y-4)^2+y^2-4(y-4)+4y-24=0...(*)
the discriminant should be 0
Now, consider the circle x²+y²-4x+4y-24+k(x-y+4)=0 and the L
sub (1) into the circle x²+y²-4x+4y-24+k(x-y+4)=0
that is
(y-4)^2+y^2-4(y-4)+4y-24+k(y-4-y+4)=0
(y-4)^2+y^2-4(y-4)+4y-24=0
that is equal to (*)
Since we know that the discriminant of (*) is 0, there is only one intersection
point of the circle x²+y²-4x+4y-24+k(x-y+4)=0 and L
So, we conclude that the circle x²+y²-4x+4y-24+k(x-y+4)=0 touches L

👍 0 👎 0