共50条信息
已知圆\(M\):\({x}^{2}+{y}^{2}-2ay-0(a > 0) \)截直线\(x+y=0\)所得线段的长度是\(2 \sqrt{2} \),则圆\(M\)与圆\(N\):\((x-1{)}^{2}+(y-1{)}^{2}=1 \)的位置关系是
已知\(⊙C:{x}^{2}+{y}^{2}=1 \),直线\(l:x+y-1=0 \),则\(l\)被\(⊙C \)所截得的弦长为( )
直线\(l:x+y+a=0\)被圆\(C:{{x}^{2}}+{{y}^{2}}=3\)截得的弦长为\(\sqrt{3}\),则\(a=\) \((\) \()\)
直线\(3x-4y-4=0\)被圆\({{(x-3)}^{2}}+{{y}^{2}}=9\)截得的弦长为( )
过点\((3,1)\)且与直线\(x-2y-3=0\)垂直的直线方程是\((\) \()\)
直线\(ax+y-5=0\)截圆\(C\):\({{x}^{2}}+{{y}^{2}}-4x-2y+1=0\)的弦长为\(4\),则\(a=(\) \()\)
进入组卷