Given:
Centre (1, 1) and radius \( \sqrt{2}\)
Let us consider the equation of a circle with centre (h, k) and
Radius r is given as (x – h)2 + (y – k)2 = r2
So, centre (h, k) = (1, 1) and radius (r) = \( \sqrt{2}\)
The equation of the circle is
(x-1)2 + (y-1)2 = \( (\sqrt{2})^2\)
x2 – 2x + 1 + y2 -2y + 1 = 2
x2 + y2 – 2x -2y = 0
The equation of the circle is x2 + y2 – 2x -2y = 0
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