Question

If the circles $x^2+y^2=9$ and $x^2+y^2-8x-6y+n^2=0,n\in Z$ have exactly two common tangents, then the number of possible values of $n$ is

Answer 1

Given circles are
$x^2+y^2=9$ and $x^2+y^2-8x-6y+n^2=0,n\in Z$
Now, the centre and radius of the circle are
$C_1=(0,0),r_1=3C_2=(4,3),r_2=\sqrt{25-n^2}$
For the square root to be defined,
$25-n^2\ge 0\Rightarrow n^2\le 25\Rightarrow n\in [-5,5]\cdots \left(1\right)$

As the given circles have exactly $2$ common tangents, so
$r_1+r_2>C_1{C}_2>|r_1-r_2|\Rightarrow 3+\sqrt{25-n^2}>\sqrt{4^2+3^2}>|3-\sqrt{25-n^2}|$

$3+\sqrt{25-n^2}>\sqrt{4^2+3^2}\Rightarrow \sqrt{25-n^2}>2$
Squaring on both sides, we get
$25-n^2>4\Rightarrow 21>n^2\Rightarrow n\in (-\sqrt{21},\sqrt{21})\cdots \left(2\right)$

$\sqrt{4^2+3^2}>|3-\sqrt{25-n^2}|\Rightarrow |3-\sqrt{25-n^2}|<5\Rightarrow -5<3-\sqrt{25-n^2}<5\Rightarrow -8<-\sqrt{25-n^2}<2\Rightarrow 8>\sqrt{25-n^2}>-2\Rightarrow 8>\sqrt{25-n^2}>0$
Squaring on both sides, we get
$64>25-n^2>0\Rightarrow 39>-n^2>-25\Rightarrow -39<n^2<25\Rightarrow 0<n^2<25\Rightarrow n\in (-5,5)\cdots \left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get
$n\in (-\sqrt{21},\sqrt{21})$
As $n\in Z$, so
$n=-4,-3,-2,-1,0,1,2,3,4$

Hence, the number of values of $n$ is $9.$