Question:
If the circles $x^{2}+y^{2}-16 x-20 y+164=r^{2}$ and $(x-4)^{2}+(y-7)^{2}=36$ intersect at two distinct points, then:
Correct Option: 4,
Solution:
Consider the equation of circles as,
$x^{2}+y^{2}-16 x-20 y+164=r^{2}$
i.e. $(x-8)^{2}+(y-10)^{2}=r^{2}$............(1)
and $(x-4)^{2}+(y-7)^{2}=36$...............(2)
Both the circles intersect each other at two distinct points.
Distance between centres
$=\sqrt{(8-4)^{2}+(10-7)^{2}}=5$
$\therefore \quad|r-6|<5<|r+6|$
$\therefore \quad$ If $|r-6|<5 \Rightarrow r \in(1,11)$..............(3)
and $|r+6|>5 \Rightarrow r \in(-\infty,-11) \cup(-1, \infty)$.....(4)
From (3) and (4),
$r \in(1,11)$