Question

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

• A. $2$
• B. $8$
• C. $9$
• D. $5$

Answer 1

Answer: (C)

$9$

$S_1:x^2+y^2=9$
$C_1\equiv (0,0)$ and $r_1=3$
$S_2:x^2+y^2-8x-6y+n^2=0$
$C_2\equiv (4,3)$ and $r_2=\sqrt{25-n^2}$

$r_2=\sqrt{25-n^2}>0$ so $-5\le n\le 5$
$\therefore$ Given $S_1$ and $S_2$ have two common tangents, $S_1$ and $S_2$ are intersecting each other.
$\therefore r_1+r_2>C_1{C}_2>|r_1-r_2|$

$C_1{C}_2>|r_1-r_2|$ is true for all $n$.

$3+\sqrt{25-n^2}>5$
$\sqrt{25-n^2}>2$
$25-n^2>4$
$n^2<21$
$(n-\sqrt{21})(n+\sqrt{21})<0$

$n\in (-\sqrt{21},\sqrt{21})$ ...(1)

and $-5\le n\le 5$ ...(2)

From (1) & (2) $n\in (-\sqrt{21},\sqrt{21}),n\in Z.$
$n=-4,-3,-2,-1,0,1,2,3,4$
$\Rightarrow$ Number of values of $n$ are $9$