If the circles
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:

1. (1) $r>11$

2. (2) $0<\mathrm{r}<1$

3. (3) $\mathrm{r}=11$

4. (4) $1<\mathrm{r}<11$

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),

$r \in(1,11)$