The equation of the circle with centre at (1, | vedclass

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(A) The given circle is $x^2 + y^2 + 2y - 3 = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 0$ and $f = 1$. The

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$x^2 + y^2 - 2x + 4y + 3 = 0$

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(A) The given circle is $x^2 + y^2 + 2y - 3 = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 0$ and $f = 1$. The centre of this circle is $(-g, -f) = (0, -1)$.The required circle has its centre at $(1, -2)$ and passes through $(0, -1)$.The radius $r$ is the distance between $(1, -2)$ and $(0, -1)$,which is $r = \sqrt{(1 - 0)^2 + (-2 - (-1))^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$,where $(h, k) = (1, -2)$.$(x - 1)^2 + (y + 2)^2 = (\sqrt{2})^2$$x^2 - 2x + 1 + y^2 + 4y + 4 = 2$$x^2 + y^2 - 2x + 4y + 3 = 0$.

The equation of the circle with centre at $(1, -2)$ and passing through the centre of the given circle $x^2 + y^2 + 2y - 3 = 0$ is

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