how do you get the equation of a circle if only given the line that goes through the centre and that x axis and y axis are tangents?

x + 2y - 6 = 0

$$x + 2y - 6 = 0\\ 2y=-x+6\\ \boxed{y=f(x)=-\frac{1}{2}x+3}\\\\ \small{\text{ r = radius of the circle }}\\ \small{\text{ $ x=f(x)=r $ }}\\ \small{\text{ $ x=-\frac{1}{2}x+3 $ }}\\ \small{\text{$\frac{1}{2}x+x=3 $}}\\ \small{\text{$\frac{3}{2}x=3 $}}\\ \small{\text{$x=\frac{2}{3}*3=2$}}\\ \small{\text{$r=2$. The center of the circle is $(x_m,\ y_m) =\ ( 2,\ 2)$ }}\\ \small{\text{The equation of the circle is: }}\\ \small{\text{ $ (x-x_m)^2+(y-y_m)^2=r^2 \qquad (x-2)^2+(y-2)^2=2^2 $ }}$$

We're looking for a point on the line that is an equal distance to both axis..........thus, we're looking for a point such that x = y

So, we have

x + 2y - 6 = 0      let y = x

x + 2x - 6  = 0

3x - 6  = 0

3x = 6

x = 2

And by the same rationale, y= 2

Thus, our center is (2,2)   and the radius = 2

Here's a graph...

GRAPH