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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 $(h, k) = (-2, 3)$ and radiu

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$x^{2}+y^{2}+4x-6y-3=0$

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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 $(h, k) = (-2, 3)$ and radius $r = 4$.Substituting these values into the formula:$(x - (-2))^{2} + (y - 3)^{2} = 4^{2}$$(x + 2)^{2} + (y - 3)^{2} = 16$Expanding the squares:$(x^{2} + 4x + 4) + (y^{2} - 6y + 9) = 16$$x^{2} + y^{2} + 4x - 6y + 13 = 16$$x^{2} + y^{2} + 4x - 6y - 3 = 0$.

Find the equation of the circle with centre $(-2, 3)$ and radius $4$.

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