问题描述: 过点A(5,0)且与圆B(x+5)^2+y^2=36相切的圆的圆心轨迹方程 1个回答 分类:数学 2014-10-01 问题解答: 我来补答 圆心C(x,y)R=√[(x-5)²+y²]圆心B(-5,0),r=5圆心距d=√[(x+5)²+y²]相切d=R+r或|R-r|d=R+r√[(x+5)²+y²]=√[(x-5)²+y²]+5√[(x+5)²+y²]-√[(x-5)²+y²]=5到(-5,0)距离减去到(5,0)距离是定值5是双曲线的一支2a=5,a=5/2c=5b²=25-25/4=75/4x²/(25/4)-y²/(75/4)=1离(-5,0)更远,是右支x²/(25/4)-y²/(75/4)=1,x>0d=|R-r|√[(x+5)²+y²]=|√[(x-5)²+y²]-5|x²+10x+25+y²=x²-10x+25+y²-10√[(x-5)²+y²]+254x-5=-2√[(x-5)²+y²]所以4x-5 展开全文阅读