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. $0<\mathrm{r}<1$

  2. $1

  3. $r>11$

  4. $r=11$


Correct Option: , 2

Solution:

$x^{2}+y^{2}-16 x-20 y+164=r^{2}$

$\mathrm{A}(8,10), \mathrm{R}_{1}=\mathrm{r}$

$(x-4)^{2}+(y-7)^{2}=36$

$\mathrm{B}(4,7), \mathrm{R}_{2}=6$

$\left|\mathrm{R}_{1}-\mathrm{R}_{2}\right|<\mathrm{AB}<\mathrm{R}_{1}+\mathrm{R}_{2}$

$\Rightarrow 1<\mathrm{r}<11$

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now