若x²+y²=4与圆x²+y²+2ay-6=0(a>0)的公共弦长为2√3,则a=…
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![若x²+y²=4与圆x²+y²+2ay-6=0(a>0)的公共弦长为2√3,则a=…](/uploads/image/z/10748373-69-3.jpg?t=%E8%8B%A5x%26%23178%3B%2By%26%23178%3B%3D4%E4%B8%8E%E5%9C%86x%26%23178%3B%2By%26%23178%3B%2B2ay%EF%BC%8D6%3D0%28a%EF%BC%9E0%29%E7%9A%84%E5%85%AC%E5%85%B1%E5%BC%A6%E9%95%BF%E4%B8%BA2%E2%88%9A3%2C%E5%88%99a%3D%E2%80%A6)
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若x²+y²=4与圆x²+y²+2ay-6=0(a>0)的公共弦长为2√3,则a=…
若x²+y²=4与圆x²+y²+2ay-6=0(a>0)的公共弦长为2√3,则a=…
若x²+y²=4与圆x²+y²+2ay-6=0(a>0)的公共弦长为2√3,则a=…
a=1.
将第二个方程配成标准式x^2+(y+a)^2=6+a^2,可知圆心在轴负半轴上,由此可得公共弦所在直线为y=1,弦长的一半=根号3.由勾股定理有(1+a)^2+3=6+a^2,解得a=1.
看完了采纳哦~~祝学习进步!