If a circle touches the axis of x at (5, 0) and passes through the point $(4, – 1),$ it also passes through the point 

Centre of the circle is on the line $x = 5$  at a distance equal to the radius of the circle from the axis of $x$

Let the coordinates of the centre be$(5, k)$ , so the equation of the circle is  

         $(x-5{)}^2+(y-k{)}^2=k^2$ 

 As it passes through the point $(4, – 1)$

$(4-5{)}^2+(-1-k{)}^2=k^2\Rightarrow k=-1$

and the equation of the circle is  

$(x-5{)}^2+(y+1{)}^2=1$

which passes through the point$(6, – 1)$

Note 

(6, – 1) is the other end of the diameters through (4, –1) of 

the circle