Two tangent circles

Examine how the circles $x^2 + y^2 = 1$ and $(x - 3)^2 + y^2 = 4$ are placed. Their centres are $(0, 0)$ and $(3, 0)$ with radii $1$ and $2$, so the distance between the centres is $3$, equal to the sum of the radii.

External tangency

When the distance between centres equals the sum of the radii, the two circles share exactly one point on the outside. The tangent point lies on the segment joining the centres, a distance $1$ from the origin toward $(3, 0)$, namely $(1, 0)$.

The algebra confirms it

Subtracting one circle equation from the other kills the quadratic terms and leaves $6x - 6 = 0$, that is $x = 1$. Putting this back gives $y^2 = 0$, so $y = 0$ arrives as a repeated root. Two shared points have merged into one: that is what tangency looks like in algebra, exactly as a vanishing discriminant signals a line tangent to a parabola.

The line $x = 1$, through the point of tangency and perpendicular to the line joining the centres, is the common tangent of the two circles. Since a tangent to a circle is perpendicular to the radius at the point of contact, one single line serves as the tangent to both.

Internal tangency

Besides external tangency there is internal tangency, where the distance between centres equals the difference of the radii and one circle touches the other from inside. Take the circle of centre $(1, 0)$ and radius $2$ against the unit circle: the centres are $1$ apart and the radii differ by $1$, so they touch from within at the single point $(-1, 0)$.

Five cases

With $d$ the distance between centres and $r_1, r_2$ the radii, the relationship falls into five cases¹.

The merit of this list is that the centres and radii settle the question before any equation is solved.

The large dot marks the point of external tangency.

1. Tangent lines to circles, Wikipedia