x^2+y^2+4x-4y-10=0,C2:x^2+y^2-4x+4y+6=0,证明这两圆相切,并求过切点的切线方程
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![x^2+y^2+4x-4y-10=0,C2:x^2+y^2-4x+4y+6=0,证明这两圆相切,并求过切点的切线方程](/uploads/image/z/3702727-55-7.jpg?t=x%5E2%2By%5E2%2B4x-4y-10%3D0%2CC2%EF%BC%9Ax%5E2%2By%5E2-4x%2B4y%2B6%3D0%2C%E8%AF%81%E6%98%8E%E8%BF%99%E4%B8%A4%E5%9C%86%E7%9B%B8%E5%88%87%2C%E5%B9%B6%E6%B1%82%E8%BF%87%E5%88%87%E7%82%B9%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B)
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x^2+y^2+4x-4y-10=0,C2:x^2+y^2-4x+4y+6=0,证明这两圆相切,并求过切点的切线方程
x^2+y^2+4x-4y-10=0,C2:x^2+y^2-4x+4y+6=0,证明这两圆相切,并求过切点的切线方程
x^2+y^2+4x-4y-10=0,C2:x^2+y^2-4x+4y+6=0,证明这两圆相切,并求过切点的切线方程
C1:①x²+y²+4x-4y-10=0,即:﹙x+2﹚²+﹙y-2﹚²=﹙3√2﹚²,
∴圆心为A﹙-2,2﹚,半径=3√2,
C2:②x²+y²-4x+4y+6=0,即﹙x-2﹚²+﹙y+2﹚²=﹙√2﹚²,
∴圆心为B﹙2,-2﹚,半径=√2,
①-②得:8x-8y-16=0,
∴③x=y+2,代入①化简得:
﹙y+1﹚²=0,∴y=-1,
代入③得:x=1,
即两圆只有一个交点,交点坐标为C﹙1,-1﹚,
∴两圆相切.
由两个圆心坐标得:AB直线方程:y=-x,
∴可以设经过切点的直线方程为:y=x+b,
将C点坐标代入得:b=-2,
∴切线方程为:y=x-2.
如果A={(x,y)|x,y+2>0},B={(x,y)|2x+3y-6>0},C={(x,y)|x-4
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