設圓心為(0,b)及b>0,並相切於原點、切線x=-0.5k及切線x=0.5k,求圓的方程。?

2016-07-14 3:47 am

回答 (1)

2016-07-14 10:10 am
Sol
設圓:x^2+(y-b)^2=r^2,r>0
(0,b)到x+0.5k=0距離=r
|0+0.5k|/1=r
r^2=0.25k^2
x^2+(y-b)^2=0.25k^2


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