For the two circles x^2+y^2=16 and x^2+y^2-2y=0, there is/are:

Question

For the two circles $x^2+y^2=16$ and $x^2+y^2-2y=0$, there is/are:

Accepted Answer

Circle 1: centre $(0,0)$, $r_1=4$. Circle 2: $x^2+(y-1)^2=1$, centre $(0,1)$, $r_2=1$.

Distance between centres $=1$. Since $|r_1-r_2|=3>1=$ distance, circle 2 lies entirely inside circle 1 without touching --- hence no common tangent.

Solutions · 1