题目内容
如图,直角坐标系中直线AB交x轴,y轴于点A(4,0)与B(0,-3),现有一半径为1的动圆的圆心位于原点处,以每秒1个单位的速度向右作平移运动,则经过______秒后动圆与直线AB相切.
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∵OA=4,OB=3,
∴AB=5,
设⊙P经过x秒后与直线AB相切,过P点作AB的垂线,垂足为Q,则PQ=1;
(1)当⊙P在直线AB的左边与直线AB相切时,AP=4-x,
由△APQ∽△ABO得,
=
,即
=
,
解得x=
;
(2)当⊙P在直线AB的右边与直线AB相切时,AP=x-4;
由△APQ∽△ABO得,
=
,即
=
,
解得x=
.
故填
或
.
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∴AB=5,
设⊙P经过x秒后与直线AB相切,过P点作AB的垂线,垂足为Q,则PQ=1;
(1)当⊙P在直线AB的左边与直线AB相切时,AP=4-x,
由△APQ∽△ABO得,
AP |
AB |
PQ |
BO |
4-x |
5 |
1 |
3 |
解得x=
7 |
3 |
(2)当⊙P在直线AB的右边与直线AB相切时,AP=x-4;
由△APQ∽△ABO得,
AP |
AB |
PQ |
BO |
x-4 |
5 |
1 |
3 |
解得x=
17 |
3 |
故填
7 |
3 |
17 |
3 |
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