求过两圆c1:x^2 y^2-4x+2y+1=0和c2:x^2+y^2-6x=0的交点且过点(2,-2)的圆的方程求详解,请仔细读题
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![求过两圆c1:x^2 y^2-4x+2y+1=0和c2:x^2+y^2-6x=0的交点且过点(2,-2)的圆的方程求详解,请仔细读题](/uploads/image/z/2085198-6-8.jpg?t=%E6%B1%82%E8%BF%87%E4%B8%A4%E5%9C%86c1%3Ax%5E2+y%5E2-4x%2B2y%2B1%3D0%E5%92%8Cc2%3Ax%5E2%2By%5E2-6x%3D0%E7%9A%84%E4%BA%A4%E7%82%B9%E4%B8%94%E8%BF%87%E7%82%B9%EF%BC%882%2C-2%EF%BC%89%E7%9A%84%E5%9C%86%E7%9A%84%E6%96%B9%E7%A8%8B%E6%B1%82%E8%AF%A6%E8%A7%A3%2C%E8%AF%B7%E4%BB%94%E7%BB%86%E8%AF%BB%E9%A2%98)
求过两圆c1:x^2 y^2-4x+2y+1=0和c2:x^2+y^2-6x=0的交点且过点(2,-2)的圆的方程求详解,请仔细读题
求过两圆c1:x^2 y^2-4x+2y+1=0和c2:x^2+y^2-6x=0的交点且过点(2,-2)的圆的方程
求详解,请仔细读题
求过两圆c1:x^2 y^2-4x+2y+1=0和c2:x^2+y^2-6x=0的交点且过点(2,-2)的圆的方程求详解,请仔细读题
公式C1+λC2=0,再把点(2,-2)代入方程C1+λC2=0得出λ=-0.75.
将λ代入C1+λC2=0得出的就是所要求的圆为:X^2+y^2-34X+8y+4=0
解由所求的圆过两圆c1:x^2 +y^2-4x+2y+1=0和c2:x^2+y^2-6x=0的交点
故设所求的圆的方程为
x^2 +y^2-4x+2y+1+t(x^2+y^2-6x)=0
又由所求的圆过点(2,-2)
即2^2 +(-2)^2-4x2+2(-2)+1+t(2^2 +(-2)^2-6x2)=0
即-3+t(-4)=0
解得t=-3/4<...
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解由所求的圆过两圆c1:x^2 +y^2-4x+2y+1=0和c2:x^2+y^2-6x=0的交点
故设所求的圆的方程为
x^2 +y^2-4x+2y+1+t(x^2+y^2-6x)=0
又由所求的圆过点(2,-2)
即2^2 +(-2)^2-4x2+2(-2)+1+t(2^2 +(-2)^2-6x2)=0
即-3+t(-4)=0
解得t=-3/4
故
所求的圆的方程为
x^2 +y^2-4x+2y+1+(-3/4)(x^2+y^2-6x)=0
即为1/4x^2 +1/4y^2-1/2x+2y+1=0
即为x^2 +y^2-2x+8y+4=0
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