动圆M与圆c1 (x+3)²+y²=9外切,且与圆c2(x-3)²+y²=1内切,则圆心M的轨迹方程!
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![动圆M与圆c1 (x+3)²+y²=9外切,且与圆c2(x-3)²+y²=1内切,则圆心M的轨迹方程!](/uploads/image/z/5261287-31-7.jpg?t=%E5%8A%A8%E5%9C%86M%E4%B8%8E%E5%9C%86c1+%EF%BC%88x%2B3%EF%BC%89%26%23178%3B%2By%26%23178%3B%3D9%E5%A4%96%E5%88%87%2C%E4%B8%94%E4%B8%8E%E5%9C%86c2%EF%BC%88x-3%EF%BC%89%26%23178%3B%2By%26%23178%3B%3D1%E5%86%85%E5%88%87%2C%E5%88%99%E5%9C%86%E5%BF%83M%E7%9A%84%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8B%21)
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动圆M与圆c1 (x+3)²+y²=9外切,且与圆c2(x-3)²+y²=1内切,则圆心M的轨迹方程!
动圆M与圆c1 (x+3)²+y²=9外切,且与圆c2(x-3)²+y²=1内切,
则圆心M的轨迹方程!
动圆M与圆c1 (x+3)²+y²=9外切,且与圆c2(x-3)²+y²=1内切,则圆心M的轨迹方程!
圆心C1(-3,0),r1=3
C2(3,0),r2=1
设M(x,y),半径是r
外切
则MC1=r+3
内切则MC2=r-1
相减
MC1-MC2=4
所以M轨迹是双曲线 ,2a=4
a=2
C1C2是焦点,c=3
则b²=3²-2²=5
因为MC1>MC2,所以是右支
所以
x²/4-y²/5=1,x>0