Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Let the equation of the required circle be $(x-h)^{2}+(y-k)^{2}=r^{2}$.
Since the radius of the circle is 5 and its centre lies on the $x$-axis, $k=0$ and $r=5$.
Now, the equation of the circle becomes $(x-h)^{2}+y^{2}=25$.
It is given that the circle passes through point $(2,3)$.
$\therefore(2-h)^{2}+3^{2}=25$
$\Rightarrow(2-h)^{2}=25-9$
$\Rightarrow(2-h)^{2}=16$
$\Rightarrow 2-h=\pm \sqrt{16}=\pm 4$
If $2-h=4$, then $h=-2$.
If $2-h=-4$, then $h=6$
When $h=-2$, the equation of the circle becomes
$(x+2)^{2}+y^{2}=25$
$x^{2}+4 x+4+y^{2}=25$
$x^{2}+y^{2}+4 x-21=0$
When h = 6, the equation of the circle becomes
$(x-6)^{2}+y^{2}=25$
$x^{2}-12 x+36+y^{2}=25$
$x^{2}+y^{2}-12 x+11=0$
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