Equation of a circle which passes through (3, 6) and touches the axes is

When circle touches both the axes, the co – ordinates of the centre and its radius are equal in their magnitude,

h = k – r

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

(3 – h)² + (6 – h)² = h²

9 + h² - 6h + 36 + h² - 12h = h²

h² - 18h + 45 = 0

h² - 15h – 3h + 45 = 0

h (h – 15) – 3 (h – 15) = 0

(h – 3) (h – 15) = 0

h = 3 or h = 15

Co – ordinates of centre are (3, 3) or (15, 15)

(x – h)² + (y – k)² = r²

Equation, having centre (3, 3)

(x – 3)² + (y – 3)² = 3²

x² - 6x + 9 + y² - 6y + 9 – 9 = 0

x² - 6x + y² - 6y + 9 = 0

Equation, having centre (15, 15)

(x – 15)² + (y – 15)² = 15²

x² - 30x + 225 + y² - 30y + 225 – 225 = 0

x² - 30x + y² - 30y + 225 = 0

Hence the equations are x² - 6x + y² - 6y + 9 = 0 or x² - 30x + y² - 30y + 225 = 0.

Option (C) is the answer.