已知圆C与直线x-y=0及x-y-4=0都相切,圆心在直线x+y=0上,则圆C的方程为()
A: (x+1)2+(y-1)2=2
B: (x-1)2+(y+1)2=2
C: (x-1)2+(y-1)2=2
D: (x+1)2+(y+1)2=2
A: (x+1)2+(y-1)2=2
B: (x-1)2+(y+1)2=2
C: (x-1)2+(y-1)2=2
D: (x+1)2+(y+1)2=2
举一反三
- 已知圆C与直线x-y=0及x-y-4=0都相切,圆心在直线x+y=0上,则圆C的方程为______ A: (x+1)2+(y-1)2=2 B: (x-1)2+(y+1)2=2 C: (x-1)2+(y-1)2=2 D: (x+1)2+(y+1)2=2 E: 以上答案均不正确
- 如下程序的运行结果是( ) intx=1,y=1;if(x==1) y=x+1;elseif(y==2) x=y+1;else y=0; A: x=1, y=2 B: x=3, y=2 C: x=3, y=0 D: x=1, y=0
- 【单选题】与曲线 y = x 2 相切,且与直线 x + 2 y + 1 = 0 垂直的直线的方程为 () A. y = 2 x - 1 B. y = 2 x + 2 C. y = 2 x - 2 D. y = 2 x + 1
- 已知齐次方程$(x-1){{y}^{''}}-x{{y}^{'}}+y=0$的通解为$Y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}$,则方程$(x-1){{y}^{''}}-x{{y}^{'}}+y={{(x-1)}^{2}}$的通解是( ) A: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-({{x}^{2}}+1)$ B: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-({{x}^{3}}+1)$ C: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}$ D: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}+1$
- 4.已知二元函数$z(x,y)$满足方程$\frac{{{\partial }^{2}}z}{\partial x\partial y}=x+y$,并且$z(x,0)=x,z(0,y)={{y}^{2}}$,则$z(x,y)=$( ) A: $\frac{1}{2}({{x}^{2}}y-x{{y}^{2}})+{{y}^{2}}+x$ B: $\frac{1}{2}({{x}^{2}}{{y}^{2}}+xy)+{{y}^{2}}+x$ C: ${{x}^{2}}{{y}^{2}}+{{y}^{2}}+x$ D: $\frac{1}{2}({{x}^{2}}y+x{{y}^{2}})+{{y}^{2}}+x$