The circle passing through (1, -2) and touching the axis of $x$ at (3, 0) also passes through the point

The system of circles touches the line $y=0$ at the point (3, 0) is given by $\left.\left\{(x-3{)}^2+y^2\right)\right\}+\lambda y=0$ As the circle passes through $(1,-2),$ we can determine $\lambda$ which gives $4+4-2\lambda =0\therefore \lambda =4$

The circle is $(x-3{)}^2+y^2+4y=0$ A simple calculation
shows that $(5, -2)$ lies on the circle.