If the circles $\mathrm{S} \equiv x^2+y^2-14 x+6 y+33=0$ and $\mathrm{S}^{\prime} \equiv x^2+$ $y^2-a^2=0(a…
If the circles $\mathrm{S} \equiv x^2+y^2-14 x+6 y+33=0$ and $\mathrm{S}^{\prime} \equiv x^2+$ $y^2-a^2=0(a \in \mathrm{~N})$ have 4 common tangents then possible number of values of $a$ is
$13$
$5$
$14$
$2$
Solution
Given, $S \equiv(x-7)^2+(y+3)^2=5^2$
So $C=(7,-3), r=5$
and $S^{\prime} \equiv x^2+y^2=a^2 \Rightarrow C^{\prime}=(0,0), r^{\prime}=a$
Since, for 4 common tangent $C C^{\prime}\gtr+r^{\prime}$
$\Rightarrow \sqrt{58} \gt 5+a \Rightarrow a \lt 7.616-5 \Rightarrow a \lt 2.616$
So, $a=2$.