JEE PYQ: Circle Question 35
Question 35 - 2019 (09 Jan Shift 2)
If the circles $x^2 + y^2 - 16x - 20y + 164 = r^2$ and $(x-4)^2 + (y-7)^2 = 36$ intersect at two distinct points, then:
(1) $r > 11$
(2) $0 < r < 1$
(3) $r = 11$
(4) $1 < r < 11$
Show Answer
Answer: (4)
Solution
$C_1(8, 10)$, $C_2(4, 7)$, $r_2 = 6$. $C_1C_2 = 5$. For two intersection points: $|r - 6| < 5 < r + 6$. So $1 < r < 11$.