The equation of the circle which touches the axes of coordinates and the line 
and whose centres lie in the first quadrant is x2 + y2 – 2cx – 2cy + c2 = 0, where c is equal to
Given that the equation of the circle which touches the axes of coordinates and the line 
and whose centres lie in the first quadrant is x2 + y2 - 2cx - 2cy + c2 = 0. We need to find the value of c.
We know that for a circle x2 + y2 + 2ax + 2by + c = 0
⇒ Centre = (- a, - b)
⇒ Radius = 
For x2 + y2 - 2cx - 2cy + c2 = 0
⇒ Centre(C) = 
⇒ C = (c,c)
⇒ Radius(r) = 
⇒ 
⇒ r = c
We have 
touching the circle.
We know that the perpendicular distance between the circle and centre is equal to the radius of the circle.
We know that the perpendicular distance between from the point (x1,y1) to the line ax + by + c = 0 is 
⇒ 
⇒ 
.
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ c2 - 7c + 6 = 0
⇒ c2 - 6c - c + 6 = 0
⇒ c(c - 6) - 1(c - 6) = 0
⇒ (c - 1)(c - 6) = 0
⇒ c - 1 = 0 or c - 6 = 0
⇒ c = 1 or c = 6
∴The correct option is (d).