已知圆(x-3)^2+(y+4)^2=4和直线y=kx相交于点P,Q,则OP*OQ的值为?是21,我看错了。那下面那个式子是怎么得出的啊?
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/07 10:48:58
![已知圆(x-3)^2+(y+4)^2=4和直线y=kx相交于点P,Q,则OP*OQ的值为?是21,我看错了。那下面那个式子是怎么得出的啊?](/uploads/image/z/5907759-15-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%86%EF%BC%88x-3%EF%BC%89%5E2%2B%28y%2B4%29%5E2%3D4%E5%92%8C%E7%9B%B4%E7%BA%BFy%3Dkx%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9P%2CQ%2C%E5%88%99OP%2AOQ%E7%9A%84%E5%80%BC%E4%B8%BA%3F%E6%98%AF21%EF%BC%8C%E6%88%91%E7%9C%8B%E9%94%99%E4%BA%86%E3%80%82%E9%82%A3%E4%B8%8B%E9%9D%A2%E9%82%A3%E4%B8%AA%E5%BC%8F%E5%AD%90%E6%98%AF%E6%80%8E%E4%B9%88%E5%BE%97%E5%87%BA%E7%9A%84%E5%95%8A%EF%BC%9F)
已知圆(x-3)^2+(y+4)^2=4和直线y=kx相交于点P,Q,则OP*OQ的值为?是21,我看错了。那下面那个式子是怎么得出的啊?
已知圆(x-3)^2+(y+4)^2=4和直线y=kx相交于点P,Q,则OP*OQ的值为?
是21,我看错了。
那下面那个式子是怎么得出的啊?
已知圆(x-3)^2+(y+4)^2=4和直线y=kx相交于点P,Q,则OP*OQ的值为?是21,我看错了。那下面那个式子是怎么得出的啊?
应该是21吧.设圆心为K,半径为r,则OP*OQ=OK^2-r^2=25-4=21
设OK交圆于M N两点
根据切割线定理知道OP*OQ=OM*ON.而OM=OK-r,ON=OK+r.所以解决
圆(x-3)^2+(y+4)^2=4和直线y=kx相交于点P,Q
y=kx
(x-3)^2+(y+4)^2=4
(x-3)^2+(KX+4)^2=4
(1+k^2)x^2+(8k-6)x+21=0
x(P)*x(Q)=21/(1+k^2)
y(P)=k*x(P),y(Q)=k*x(Q)
OP^2=[x(P)]^2+[y(P)]^...
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圆(x-3)^2+(y+4)^2=4和直线y=kx相交于点P,Q
y=kx
(x-3)^2+(y+4)^2=4
(x-3)^2+(KX+4)^2=4
(1+k^2)x^2+(8k-6)x+21=0
x(P)*x(Q)=21/(1+k^2)
y(P)=k*x(P),y(Q)=k*x(Q)
OP^2=[x(P)]^2+[y(P)]^2=(1+k^2)*[x(P)]^2
OQ^2=[x(Q)]^2+[y(Q)]^2=(1+k^2)*[x(Q)]^2
(OP*OQ)^2=[x(P)*x(Q)]^2*(1+k^2)^2=[21/(1+k^2)]^2*(1+k^2)^2=21^2
OP*OQ=21
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