Let the first circle be $S_1: x^2+y^2=4$. Its center is $C_1=(0,0)$ and its radius is $R_1=2$. Let the second circle be $S_2: x^2+y^2-6x-8y=24$. Its center is $C_2=(-(-6)/2, -(-8)/2) = (3,4)$. Its radius is $R_2 = \sqrt{3^2+4^2 - (-24)} = \sqrt{9+16+24} = \sqrt{49}=7$. The distance between the centers $d = C_1C_2 = \sqrt{(3-0)^2+(4-0)^2} = \sqrt{9+16} = \sqrt{25}=5$. Now, we compare $d$ with $R_1+R_2$ and $|R_1-R_2|$. $R_1+R_2 = 2+7 = 9$. $|R_1-R_2| = |2-7| = 5$. Since $d = |R_1-R_2|$ (i.e., $5=5$), the two circles touch each other internally. When two circles touch internally, they have exactly one common tangent.