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?
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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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