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JYSim98 2015-10-6 09:46 AM

圆与球

求圆心在x轴上,且经过二已知圆 x²+y²-2x-14y+25=0 , x²+y²+2x-2y-3=0 的交点的圆之方程式。
Ans:3x²+3y²+8x-23=0
本人做到 x²+y²-14x-11=0……唉,求解释



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Accumulators 2015-10-13 10:52 AM

You probably simply put y = 0 for the equation of straight line passing through the intersection and took that as the centre, which is not correct !

Try to find the real mid point of the chord formed by the intersection and the correct centre (which should lie on the line connecting the centres of the two circles)



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top11 2015-10-14 01:30 AM

[quote]原帖由 [i]JYSim98[/i] 於 2015-10-6 09:46 AM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=427277632&ptid=25163828][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
求圆心在x轴上,且经过二已知圆 x²+y²-2x-14y+25=0 , x²+y²+2x-2y-3=0 的交点的圆之方程式。
Ans:3x²+3y²+8x-23=0
本人做到 x²+y²-14x-11 ... [/quote]
[color=#4b0082]使用圓族 (family of circles) 的概念即成。[/color]

[color=#4b0082]令所求的圓為[/color]

[color=#0000ff]x² + y² - 2x - 14y + 25[/color][color=#4b0082] + k[/color][color=#8b0000](x² + y² + 2x - 2y - 3)[/color][color=#4b0082] = 0[/color]

[color=#4b0082](1 + k)x² + (1 + k)y² + (2k - 2)x - (2k + 14)y + 25 - 3k = 0[/color]

[color=#4b0082]圓心 = ( 1 - k , k + 7 )[/color]

[color=#4b0082]由於圓心在 x-軸上,因此 k + 7 = 0,即 k = -7。[/color]

[color=#4b0082]所求的圓是(1 - 7)x² + (1 - 7)y² + (-14 - 2)x - (-14 + 14)y + 25 + 21 = 0[/color]

[color=#4b0082]-6x² - 6y² - 16x + 46 = 0[/color]

[color=#4b0082]3x² + 3y² + 8x - 23 = 0[/color]

[color=#ff0000]✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀[/color]

(1)
[color=#0000ff]其實你就讀的是什麼課程?[/color]

(2)
[color=#008000]所謂圓族,就是指所有通過指定某兩圓交點的所有圓。[/color]
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