已知动圆P与圆M:(x+2根号6⼀3)^2+y^2=16相切,且经过点N(2根号6⼀3,0),试求动圆的圆

试求动圆的圆心P的轨迹C的方程
2025-01-07 08:57:55
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回答1:

解:设动圆圆心(x,y),半径为r。则
(1)外切时。(x+2√6/3)²+y²=(4+r)² (x-2√6/3)²+y²=r²∴相减得,8√6/3*x=16+8r∴r=√6/3*x-2
∴(x-2√6/3)²+y²=(√6/3*x-2)²∴1/3*x²+y²=4/3∴x²/4+y²/(3/4)=1
(2)内切时。(x+2√6/3)²+y²=(4-r)² (x-2√6/3)²+y²=r²∴相减得,8√6/3*x=16-8r∴r=-√6/3*x+2
∴(x-2√6/3)²+y²=(-√6/3*x+2)²∴1/3*x²+y²=4/3∴x²/4+y²/(3/4)=1
∴综上,动圆的圆心P的轨迹C的方程:x²/4+y²/(3/4)=1