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: , 2
Solution:
$x^{2}+y^{2}-16 x-20 y+164=r^{2}$
$\mathrm{A}(8,10), \mathrm{R}_{1}=\mathrm{r}$
$(x-4)^{2}+(y-7)^{2}=36$
$\mathrm{B}(4,7), \mathrm{R}_{2}=6$
$\left|\mathrm{R}_{1}-\mathrm{R}_{2}\right|<\mathrm{AB}<\mathrm{R}_{1}+\mathrm{R}_{2}$
$\Rightarrow 1<\mathrm{r}<11$
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