assume L: y = kx + b
because of P, we have k + b = 1
L and L1 intersects at y = kx + b = x + 3 or x = (3-b)/(k-1), y = (3-b)/(k-1) +3
L and L2 intersects at y = kx + b = 6-2x or x = (6-b)/(k+2) and y = 6 - 2(6-b)/(k+2)
now you have A and B's coordinates
AP = BP
you get two equations with two variables k and b
k = -1/3 a = 4/3
y = -1/3 x + 4/3
思路：1、设方程Y=ax+b;
分别与L1、L2联立两个方程组解出两个交点：C、D
立方程CP=DP解出a、b得方程
2、算出原点o到直线L的距离d，由直角三角形得：d的平方+弦长一半的平方=半径的平方
知道圆心、半径就可以写出圆的方程了
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