Sign In
JEE Advance - Mathematics (1989 - No. 20)
If the two circles $${(x - 1)^2} + {(y - 3)^2} = {r^2}$$ and $${x^2} + {y^2} - 8x + 2y + 8 = 0$$ intersect in two distinct points, then
2 < r < 8
r < 2
r = 2
r > 2
Comments (0)
Login To Comment
Advertisement
Allow javascript to properly load this page