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 =r2
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 + y2 = 25.
It is given that the circle passes through point (2, 3)
(2 – h)2+ 32 = 2
(2 – h)2 = 25-9
(2 – h)2 = 16
2 –h =
= 
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 + y2 = 25
⇒ x2 -12x + 36 + y2 = 25
⇒ x2 + y2 -4x + 21 =0
When h= 6, the equation of the circle becomes
(x – 6)2 + y2 = 25
⇒ x2 -12x + 36 + y2 = 25
⇒ x2 + y2 -12x + 11 =0
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