Centre of the circle is C (3, 1). Let the circle touch the line 8x – 15y + 25 = 0 at point M.
CM = radius (r) CM = Length of perpendicular from centre C(3, 1) on the line 8x – 15y + 25 = 0
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)2 + (y – k)2 = r2
Here, h = 3, k = 1 and r = 2
The required equation of the circle is
⇒ (x – 3)2 + (y – 1)2 = 22
⇒ x2 – 6x + 9 + y2 – 2y + 1 = 4
⇒ x2 + y2 – 6x – 2y + 10 – 4 = 0
⇒ x2 + y2 – 6x – 2y + 6 = 0
