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).