Question -
Answer -
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the radius of the circle is 5 and itscentre 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 thecircle passes through point (2, 3).
When h = –2, the equationof the circle becomes
(x + 2)2 + y2 = 25
x2 + 4x + 4+ y2 = 25
x2 + y2 + 4x –21 = 0
When h = 6, the equation ofthe circle becomes
(x – 6)2 + y2 = 25
x2 – 12x +36+ y2 = 25
x2 + y2 – 12x +11 = 0