Question

There are two circles whose equations are $x^2+y^2=4$ and $x^2+y^2-24x-10y+a^2=0,aϵZ$. If the two circle have exactly two common tangents then the number of possible values of $a$ is

• A. $2$
• B. $8$
• C. $13$
• D. none of these

Answer 1

The correct option is C $13$

Equation of circles are $x^2+y^2=4$ ...(1)
and, $x^2+y^2-24x-10y+a^2=0$ ...(2)
Center of (1) is $C_1\equiv (0,0)$ and radius $r_1=2$
Center of $C_2\equiv (12,5)$ and radius $r_2=\sqrt{169-a^2}$
$d=$ distance between centers $=C_1{C}_2=\sqrt{144+25}=13$
If the two circles have exactly two common tangents, then
$169-a^2>0$ and $r_1+r_2>d$
$\Rightarrow (a-13)(a+13)<0$ and $2+\sqrt{169-a^2}>13$
$\Rightarrow -13<a<13$ and $169-a^2>121$
$\Rightarrow -13<a<13$ and $a^2-48<0$
$\Rightarrow -13<a<13$ and $-\sqrt{48}<a<\sqrt{48}$
$\Rightarrow -\sqrt{48}<a<\sqrt{48}$
Since $a$ is an integer
$\therefore a\equiv -6,-5,-4,...,4,5,6$
Therefore the number of possible values of $a$ is $13$