Find the equation of the circle passing through ( 1 , - 2 ) and touching the x - axis at ( 3 , 0 ) .

Let required circle equation is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Given that circle touching the $x$axis at $\left(3, 0\right)$.

So $x$ axis is a tangent to the given circle i.e perpendicular distance from centre to tangent must be equal to radius.

$\Rightarrow k=r$

${\left(x-h\right)}^2+{\left(y-k\right)}^2=k^2$

It is passing through $\left(3, 0\right) \& \left(1, - 2\right)$

$\Rightarrow 9+h^2-6h =0$

$\Rightarrow {\left(h-3\right)}^2=0 \Rightarrow h=3$

${\left(1-3\right)}^2+{\left(-2-k\right)}^2=k^2$

$\Rightarrow 4+4+k^2+4k=k^2$

$\Rightarrow 4k+8=0$

$\Rightarrow k =r = -2$

So circle equation is ${\left(x-3\right)}^2+{\left(y+2\right)}^2=4$

$\therefore x^2+y^2-6x+4y+9=0$