已知圆C1:x²+y²-4x-2y-5=0与圆C2:x²+y²-6x-y-9=0相交,相交弦所在直线方程为2x-y+4=0,在平面找一点P,过P引两圆切线并使它们长都为6√2,求P坐标.设P(x,y) 则2x-y+4=0 x²+y²-6x-y-9=(6
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![已知圆C1:x²+y²-4x-2y-5=0与圆C2:x²+y²-6x-y-9=0相交,相交弦所在直线方程为2x-y+4=0,在平面找一点P,过P引两圆切线并使它们长都为6√2,求P坐标.设P(x,y) 则2x-y+4=0 x²+y²-6x-y-9=(6](/uploads/image/z/2640309-69-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%86C1%EF%BC%9Ax%26%23178%3B%2By%26%23178%3B-4x-2y-5%3D0%E4%B8%8E%E5%9C%86C2%EF%BC%9Ax%26%23178%3B%2By%26%23178%3B-6x-y-9%3D0%E7%9B%B8%E4%BA%A4%2C%E7%9B%B8%E4%BA%A4%E5%BC%A6%E6%89%80%E5%9C%A8%E7%9B%B4%E7%BA%BF%E6%96%B9%E7%A8%8B%E4%B8%BA2x-y%2B4%3D0%2C%E5%9C%A8%E5%B9%B3%E9%9D%A2%E6%89%BE%E4%B8%80%E7%82%B9P%2C%E8%BF%87P%E5%BC%95%E4%B8%A4%E5%9C%86%E5%88%87%E7%BA%BF%E5%B9%B6%E4%BD%BF%E5%AE%83%E4%BB%AC%E9%95%BF%E9%83%BD%E4%B8%BA6%E2%88%9A2%2C%E6%B1%82P%E5%9D%90%E6%A0%87.%E8%AE%BEP%EF%BC%88x%2Cy%29+%E5%88%992x-y%2B4%3D0+x%26%23178%3B%2By%26%23178%3B-6x-y-9%3D%EF%BC%886)
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已知圆C1:x²+y²-4x-2y-5=0与圆C2:x²+y²-6x-y-9=0相交,相交弦所在直线方程为2x-y+4=0,在平面找一点P,过P引两圆切线并使它们长都为6√2,求P坐标.设P(x,y) 则2x-y+4=0 x²+y²-6x-y-9=(6
已知圆C1:x²+y²-4x-2y-5=0与圆C2:x²+y²-6x-y-9=0相交,相交弦所在直线方程为2x-y+4=0,在平面找一点P,过P引两圆切线并使它们长都为6√2,求P坐标.
设P(x,y) 则2x-y+4=0 x²+y²-6x-y-9=(6√2)² 解得P(3,10)或P(-23/5,-26/5)
已知圆C1:x²+y²-4x-2y-5=0与圆C2:x²+y²-6x-y-9=0相交,相交弦所在直线方程为2x-y+4=0,在平面找一点P,过P引两圆切线并使它们长都为6√2,求P坐标.设P(x,y) 则2x-y+4=0 x²+y²-6x-y-9=(6
由题意知,点P必在相交弦上,故有2x-y+4=0,由切点,圆心C2及P构成直角三角形,由勾股定理,化简得x²+y²-6x-y-9=(6√2)²
在直线2x-y+4=0, 且距离C2:x²+y²-6x-y-9=0为6√2,即p到c2的切点的距离为6√2,可以确定唯一的一点p