两圆C1:X^2+Y^2+4y=0,C2:X^2+Y^2+2(a-1)x+2y+a^2=0在交点处的切线互相垂直,那么实数a的值?
问题描述:
两圆C1:X^2+Y^2+4y=0,C2:X^2+Y^2+2(a-1)x+2y+a^2=0在交点处的切线互相垂直,那么实数a的值?
答
C1:X^2+Y^2+4y=0,C2:X^2+Y^2+2(a-1)x+2y+a^2=0
圆心C1(0,-2)半径R1^2=4
圆心C2(1-a,-1)半径R2^2=2-2a>0,a