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 = $\sqrt{(2-h)^2+(3-0)^2}$ = $\sqrt{h^2+4-4h+9}$ = $\sqrt{h^2-4h+13}$ But radius of circle = 5 ∴ $\sqrt{h^2-4h+13}$ = 5 ⇒ $h^2$ – 4h + 13 = 25 ⇒ $h^2$ – 4h – 12 = 0 ⇒ (h – 6)(h + 2) = 0 ⇒ h = 6 or h = –2 When h = 6 Equation of circle is $(x-6)^2+(y-0)^2$ = $(5)^2$ ⇒ $x^2$ + 36 – 12x + $y^2$ = 25 ⇒ $x^2+y^2$ – 12x + 11 = 0 When h = –2 Equation of circle is $(x+2)^2+(y-0)^2$ = $(5)^2$ ⇒ $x^2$ + 4 + 4x + $y^2$ = 25 ⇒ $x^2+y^2$ + 4x – 21 = 0.