1,设圆心O(a,a-1)半径r是O到直线4x+3y+14=0的距离,化简得r=abs(7a+11)/5abs是绝对值又弦心距是O到直线3x+4y+10的距离,化简得d=abs(7a+6)/5由勾股定理可知弦长l=2×(r^2-d^2)^(1/2)=6所以r^2-d^2=9,解得a=2,所以圆心O(2,1)r=abs(7a+11)/5=5圆的方程是(x-2)^2+(y-1)^2=252,弦心距d为圆心O(2,-1)到直线y=x-1的距离化简得d=2^(1/2)由l=2×(r^2-d^2)^(1/2)得r=2所以圆的方程是(x-2)^2+(y+1)^2=43,联立方程{x^2+y
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