Question

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 :

• A. $1<r<11$
• B. $0<r<1$
• C. $r=11$
• D. $r>11$

Answer 1

The correct option is A $1<r<11$

$x^2+y^2-16x-20y+164=r^2$
$\Rightarrow (x-8{)}^2+(y-10{)}^2=r^2$
$(x-4{)}^2+(y-7{)}^2=36=6^2$

$\therefore$ The distance between their centres is,
$C_1{C}_2=\sqrt{(8-4{)}^2+(10-7{)}^2}$
$C_1{C}_2=5$
As we know,
$|r-6|<C_1{C}_2<r+6$
$|r-6|<5<|r+6|$

If $|r-6|<5$
$\Rightarrow r-6<5\text{and}r-6>-5$
$\Rightarrow r<11\text{and}r>1$
$\therefore r\in (1,11)...\left(1\right)$

and $|r+6|>5$
$\Rightarrow r+6>5\text{and}r+6<-5$
$\therefore r\in (-\infty ,-11)\cup (-1,\infty )...\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,
$r\in (1,11)$