Circle tangent to the x-axis

Consider the circle $(x - 2)^2 + (y - 2)^2 = 4$ with centre $(2, 2)$ and radius $2$. The distance from the centre to the $x$-axis is the centre's $y$-coordinate, which equals the radius.

The condition for tangency

When the distance from the centre to a line equals the radius, the circle is tangent to that line. So this circle touches the $x$-axis at the point directly below the centre, $(2, 0)$.

The algebra agrees

The $x$-axis is $y = 0$, and substituting gives $(x - 2)^2 = 0$, whose solution is a repeated root: exactly one shared point. Distance or equation, both arrive at the same tangency.

The point of contact lies directly below the centre because a tangent is perpendicular to the radius at that point. A radius perpendicular to the $x$-axis points straight down, so dropping the radius from the centre lands on the tangent point.

Tangent to both axes

Since this circle has centre $x$-coordinate $2$ and radius $2$, it is tangent to the $y$-axis as well: substituting $x = 0$ gives $(y - 2)^2 = 0$, so it touches at $(0, 2)$.

A circle tangent to both coordinate axes has its centre at $(\pm r, \pm r)$, that is, on the line $y = x$ or $y = -x$. In the first quadrant such a circle is $(x - r)^2 + (y - r)^2 = r^2$, and as $r$ grows its centre slides out along $y = x$ while the circle stays wedged snugly into the corner.

Touching both axes, this circle sits entirely inside the first quadrant: both $x$ and $y$ stay between $0$ and $4$ and never stray negative. Circles pressed into a corner like this turn up whenever inscribed circles or the packing of pipes is at issue.

The same rule for any line

Read the condition as distance from the centre equals radius, and it applies to any line, not just the axes.

The large dots mark the tangent point and the centre.