$(r_1,o_1)$ and $(r_2,o_2)$

$|r_1 - r_2|

$d_{o_1,o_2}=r_1+r_2$ (two circles touch each other externally)

$d_{o_1,o_2} = |r_1 - r_2|$ (two circles touch each other internally)

$x=a+r_1 \cos \theta$

$y=b+r_1 \sin \theta$

$x=c+r_2 \cos \phi$

$y=d+r_2 \sin \phi$

$a+r_1 \cos \theta = c+r_2 \cos \phi$

$b + r_1 \sin \theta=d+r_2 \sin \phi$

$r_1 \cos \theta = (c-a)+r_2 \cos \phi,$
$r_1 \sin \theta = (d-b) + r_2 \sin \phi$

$r_1^2 \cos^2 \theta + r_1^2 \sin^2 \theta = (c-a)^2 + r_2^2 \cos^2 \phi + 2 (c-a)r_2 \cos \phi + (d-b)^2 + r_2^2 \sin^2 \phi + 2(d-b) r_2 \sin \phi$

$r_1^2 = (c-a)^2 + (d-b)^2 + r_2^2 +2r_2 d_{o_1,o_2} \biggl(\frac{(c-a)}{d_{o_1,o_2}} \cos \phi + \frac{(d-b)}{d_{o_1,o_2}}\sin\phi \biggl)$

$r_1^2 = r_2^2+d_{o_1,o_2}^2 + 2r^2 d_{o_1,o_2} (\cos \alpha \cos \phi + \sin \alpha \sin \phi)$

$r_1^2 = r_2^2 + d_{o_1,o_2}^2 + 2r_2 d_{o_1,o_2} \cos (\phi - \alpha)$

$\cos (\phi - \alpha)=\frac{r_1^2 - (r_2^2 + d_{o_1,o_2}^2)}{2r_2 d_{o_1,o_2}}$

$\Rightarrow \phi - \alpha = \pi$

$\Rightarrow \phi = \alpha + \pi$

$r_1^2 = r_2^2 + d_{o_1,o_2}^2 - 2r_2 d_{o_1,o_2}$

$\Rightarrow r_1^2 = (r_1 - d_{o_1,o_2})^2$

$\Rightarrow r_1 = r_2 - d_{o_1,o_2}$, or $r_1 = d_{o_1,o_2}-r_2$

$d_{o_1,o_2}=r_1+r_2$

$\cos(\phi - \alpha)=\frac{r_1^2 - (r_2^2 + d_{o_1,o_2}^2)}{2r_2 d_{o_1,o_2}}$

$=-1$

$d_{o_1,o_2} = |r_1 - r_2|$

$\left|\frac{r_1^2 - (r_2^2 + d_{o_1,o_2}^2)}{2r_2 d_{o_1,o_2}}\right|<1$ (two circles intersect at exactly 2 points)

$\left|\frac{r_1^2 - (r_2^2 + d_{o_1,o_2}^2)}{2r_2 d_{o_1,o_2}}\right|=1$ (two circles touch each other)

$\left|\frac{r_1^2 - (r_2^2 + d_{o_1,o_2}^2)}{2r_2 d_{o_1,o_2}}\right|>1$ (two circles donot intersect)

$\$
$|r_1 - r_2| < d_{o_1,o_2} < r_1 + r_2$

$cos \alpha = \frac{(c-a)}{d_{o_1,o_2}}=1$

$\sin \alpha = 0 \Rightarrow \alpha = 0$

$x_1 = c + r_2 \cos \phi_1$

$y_1 = d + r_2 \sin \phi_1$

$\cos(\phi - \alpha) = \frac{r_1^2-(r_2^2 + d_{o_1,o_2}^2)}{2r_2 d_{o_1,o_2}}$

$=\frac{3^2 - (3^2 + 5^2)}{2 \times 3 \times 5} = -\frac{5}{6}$

$x_2 = c + r_2 \cos \phi_2$

$y_2 = d + r_2 \sin \phi_2$

$\cos \phi = -\frac{5}{6}$

$\phi_1 = \cos^{-1}[\frac{-5}{6}] \in [0, \pi]$

$\phi_2 = 2\pi - \cos^{-1} (-5/6)$

Thank you