The number of common tangents to the circles $x^{2}+y^{2}=4$ and $x^{2}+y^{2}-6 x-8 y-24=0$ is,

We have, Equation of circle is $x^{2}+y^{2}=4$. Its centre is $C_{1}(0,0)$ and radius $\left(r_{1}\right)$ is 2 . Equation of another circle is $$ x^{2}+y^{2}-6 x-8 y-24=0 $$ i.e. $(x-3)^{2}+(y-4)^{2}=\langle 7)^{2}$. Its centre is $C 2(3,4)$ and radius $\left\langle r_{2}\right\}$ is 7 . Nos, $C_{1} C_{g}=\sqrt{(3-0)^{2}+(4-0)^{2}}=5$ and $\left|n_{1}-r_{2}\right|=|2-7|=5$ $$ \therefore \quad C_{1} C_{2}=\left|r_{1}-r_{2}\right| $$ So, both circles touch each other internally.