Find the equation of the circle with radius $5$ whose centre lies on $x$-axis and passes through the point $(2,3)$

So, let the center be $$(h,0)$$
Equation of circle is
$$(x-h)^2+y^2=25$$       ....(1)
Since, it passes through $$(2,3)$$
$$\Rightarrow (2-h)^2+3^2=25$$
$$\Rightarrow (2-h)^2=16$$
$$\Rightarrow 2-h=\pm 4$$
$$\Rightarrow h=-2, h=6$$

When $$h=-2$$, the equation of circle is 
$$(x+2)^2+y^2=25$$
$$\Rightarrow x^2+4x+4+y^2=25$$
$$\Rightarrow x^2+y^2+4x-21=0$$

When $$h=6$$, the equation of circle is 
$$(x-6)^2+y^2=25$$
$$\Rightarrow x^2-12x+36+y^2=25$$
$$\Rightarrow x^2+y^2-12x+11=0$$