The equation of the circle which touches the x -axis, passes through the point (1,1) and whose centre lies on the line x+y=3 in the first quadrant is

Step-by-step solution

Let the centre of the circle be (h, k) . Since the circle touches the x -axis, the radius r is equal to |k| . As the circle is in the first quadrant, k > 0 , so r = k . The centre (h, k) lies on the line x + y = 3 , so h + k = 3 , which implies h = 3 - k . The equation of the circle is (x - h)^2 + (y - k)^2 = k^2 . Substituting h = 3 - k , we get (x - (3 - k))^2 + (y - k)^2 = k^2 . The circle passes through (1, 1) , so (1 - (3 - k))^2 + (1 - k)^2 = k^2 . Simplifying the equation: (k - 2)^2 + (1 - k)^2 = k^2 . k^2 -