过点\(P\)作圆\(C\)的切线，切点为\(Q\)，再过\(P\)作圆\(C':{(x + 2)^2} + {(y + 2)^2} = 1\)的切线，切点为

“微信扫一扫”进入题库练习及模拟考试

已知圆\(C:{x^2} + {y^2} - 4x - 2y + m = 0\)与直线\(l:3x - 4y - 7 = 0\)相交于\(M,N\)两点且\(|MN|=2\sqrt {3}\)；

过点\(P\)作圆\(C\)的切线，切点为\(Q\)，再过\(P\)作圆\(C':{(x + 2)^2} + {(y + 2)^2} = 1\)的切线，切点为\(R\)，若\(|PQ| = |PR|\)，求\(|OP|\)的最小值(其中\(O\)为坐标原点).

参考答案：\(\frac{3}{5}\)