Find the Equation of the Circle which passes through 2 points on x - axis which are at a distance of 4 units from origin and whose radius is 5 units

The general equation of the circle is,
$x^2+y^2+2gx+2fy+c=0$... (1)
The two points on the x-axis are at the distance of 4 units from the orgin.
So the points can be (4, 0) and (-4, 0).
So (1) passes through these 2 points.
$$\begin{gathered} \left(4,0\right)\Rightarrow 4^2+0^2+2g. 4+ 2f.0+c=0\Rightarrow 16+8g+c=0 ... \left(2\right) \\ \left(-4,0\right)\Rightarrow {\left(-4\right)}^2+0^2+2g. \left(-4\right)+ 2f.0+c=0\Rightarrow 16-8g+c=0 ... \left(3\right) \\ Adding \left(2\right) and \left(3\right) we get, \\ 32+2c=0 \\ c=\frac{-32}{2}=-16 \\ From \left(2\right), \\ 16+8g -16=0 \\ 8g=0 \\ g=0 \\ Given tht radius = 5 \\ \Rightarrow \sqrt{g^2+f^2-c}=5 \\ \Rightarrow \sqrt{0^2+f^2-\left(-16\right)}=5 \\ \Rightarrow f^2+16=25 \\ \Rightarrow f^2=9 \\ \Rightarrow f=\pm 3 \\ Substituting g, f and c values in \left(1\right), \\ {x}^2+y^2+2x. 0+2y. \left(\pm 3\right)-16=0 \\ \Rightarrow x^2+y^2\pm 6y-16=0 \end{gathered}$$