Question

矩形ABCD中,AB=3,BC=4,P是边AD上一点,过三点A,B,C作圆O（3）当CD与圆O相切时,求BC被圆O截得的

Answer 1

记圆与BC的交点为M,过圆心O做ON⊥BM,过圆心O做EF⊥AB,交AB于E,交CD于F

显然EF为AB的垂直平分线,EF=AB=3,OF⊥CD,即F为圆与CD的切点

OF^2=OA^2=AE^2+OE^2

(EF-OE)^2=AE^2+OE^2

(4-OE)^2=1.5^2+OE^2

OE=55/32

显然EOND为矩形,BN＝OE

ON为BM的垂直平分线,BM＝2BN＝2*55/32＝55/16