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 }= r^{2}

So, centre (h, k) = (\( \dfrac{1}{2}\), \( \dfrac{1}{4}\)) and radius (r) =\( \dfrac{1}{12}\)

The equation of the circle is

(x – \( \dfrac{1}{2}\))^{2} + (y – \( \dfrac{1}{4}\))^{2} = (\( \dfrac{1}{12}\))^{2}

x^{2} – x + \( \dfrac{1}{4}\) + y^{2} – \( \dfrac{y}{2}\) + \( \dfrac{1}{16}\) = \( \dfrac{1}{144}\)

x^{2} – x + \( \dfrac{1}{4}\) + y^{2} – \( \dfrac{y}{2}\) + \( \dfrac{1}{16}\) = \( \dfrac{1}{144}\)

144x^{2} – 144x + 36 + 144y^{2} – 72y + 9 – 1 = 0

144x^{2} – 144x + 144y^{2} – 72y + 44 = 0

36x^{2} + 36x + 36y^{2} – 18y + 11 = 0

36x^{2} + 36y^{2} – 36x – 18y + 11= 0

The equation of the circle is 36x^{2} + 36y^{2} – 36x – 18y + 11= 0

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