Find the equation of the circle with centre l | vedclass

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(A) The equation of a circle with centre $(h, k)$ and radius $r$ is given by $(x-h)^{2} + (y-k)^{2} = r^{2}$.Given centre $(h, k) = \left(\frac{1}{2},

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

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(A) The equation of a circle with centre $(h, k)$ and radius $r$ is given by $(x-h)^{2} + (y-k)^{2} = r^{2}$.Given centre $(h, k) = \left(\frac{1}{2}, \frac{1}{4}\right)$ and radius $r = \frac{1}{12}$.Substituting these values into the formula:$\left(x - \frac{1}{2}\right)^{2} + \left(y - \frac{1}{4}\right)^{2} = \left(\frac{1}{12}\right)^{2}$$x^{2} - x + \frac{1}{4} + y^{2} - \frac{y}{2} + \frac{1}{16} = \frac{1}{144}$Multiplying the entire equation by $144$ to clear the denominators:$144x^{2} - 144x + 36 + 144y^{2} - 72y + 9 = 1$$144x^{2} + 144y^{2} - 144x - 72y + 45 - 1 = 0$$144x^{2} + 144y^{2} - 144x - 72y + 44 = 0$Dividing by $4$:$36x^{2} + 36y^{2} - 36x - 18y + 11 = 0$.

Find the equation of the circle with centre $\left(\frac{1}{2}, \frac{1}{4}\right)$ and radius $\frac{1}{12}$.

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