X^2+Y^2就是到原点(0,0)距离
X^2+Y^2-6X-4Y+12=0
(x-3)^2+(y-2)^2=1
参数方程:x=3+cost,y=2+sint
假设tanp=6/4=3/2
x^2+y^2
=4sint+6cost+14
=2√13sin(x+p)+14
14-2√13
已知P(X,Y)为圆X^2+Y^2-6X-4Y+12=0上的动点,求X^2+Y^2的取值范围
已知P(X,Y)为圆X^2+Y^2-6X-4Y+12=0上的动点,求X^2+Y^2的取值范围
说个方法也行
说个方法也行
数学人气:413 ℃时间:2020-04-29 06:29:44
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