F.6 math
2013-09-26 4:56 am
the centre of the circle x^2+y^2-kx-(k+2)y+9=0 lies on x+y-7=0,where k is a constant.find the value of k
回答 (2)
2013-09-26 5:11 am
✔ 最佳答案
the centre of the circle x^2+y^2－kx－(k+2)y+9=0 lies on x+y－7=0，where kis a constant.find the value of k
Sol
x^2+y^2－kx－(k+2)y+9=0
(x^2－kx)+[y^2－(k+2)y]=－9
(x^2－kx+k^2/4)+[y^2－(k+2)y+(k+2)^2/4]=－9+k^2/4+(k+2)^2/4
(x－k/2)^2+[y－(k+2)/2]^2=－9+k^2/4+(k+2)^2/4
圓心(k/2，k/2+1)
k/2+k/2+1－7=0
k=6
－9+36/4+64/4>0
2013-09-26 5:36 am
x^2+y^2+Dx+Ey+F=0
Centre=(-D/2,-E/2)
x^2+y^2-kx-(k+2)y+9=0
Centre={ -(-k)/2,-[-(k+2)]/2 } =[k/2,(k+2)/2]
x+y-7=0
k/2+(k+2)/2-7=0
k+k+2-14=0
2k=12
k=6
Centre=(-D/2,-E/2)
x^2+y^2-kx-(k+2)y+9=0
Centre={ -(-k)/2,-[-(k+2)]/2 } =[k/2,(k+2)/2]
x+y-7=0
k/2+(k+2)/2-7=0
k+k+2-14=0
2k=12
k=6
收錄日期: 2021-04-20 14:24:14
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130925000051KK00220