Since the centre of the circle lies on x-axis, the co-ordinates of centre are (h, 0).
Now the circle passes through the point (2, 3).
∴ Radius of circle
= √(2−h)2+(3−0)2 = √h2+4−4h+9 = √h2−4h+13
But radius of circle = 5
∴ √h2−4h+13 = 5
⇒ h2 – 4h + 13 = 25 ⇒ h2 – 4h – 12 = 0
⇒ (h – 6)(h + 2) = 0 ⇒ h = 6 or h = –2
When h = 6
Equation of circle is (x–6)2+(y–0)2 = (5)2
⇒ x2 + 36 – 12x + y2 = 25 ⇒ x2+y2 – 12x + 11 = 0
When h = –2
Equation of circle is (x+2)2+(y–0)2 = (5)2
⇒ x2 + 4 + 4x + y2 = 25 ⇒ x2+y2 + 4x – 21 = 0.