Given:
Centre (\( \dfrac{1}{2}\), \( \dfrac{1}{4}\)) and radius \( \dfrac{1}{12}\)
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) = (\( \dfrac{1}{2}\), \( \dfrac{1}{4}\)) and radius (r) =\( \dfrac{1}{12}\)
The equation of the circle is
144x2 – 144x + 36 + 144y2 – 72y + 9 – 1 = 0
144x2 – 144x + 144y2 – 72y + 44 = 0
36x2 + 36x + 36y2 – 18y + 11 = 0
36x2 + 36y2 – 36x – 18y + 11= 0
The equation of the circle is 36x2 + 36y2 – 36x – 18y + 11= 0
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