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。?
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
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
收錄日期: 2021-04-18 15:14:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160713194726AAi4MfR