Find the equation of the circle with radius 5 whose centre

Solution:

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

$(x-6)^{2}+y^{2}=25$

$x^{2}-12 x+36+y^{2}=25$

$x^{2}+y^{2}-12 x+11=0$