The gradient of the normal at the point (-2, | vedclass

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(A) The given equation of the circle is $x^2 + y^2 + 2x + 4y + 3 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g =

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$1$

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(A) The given equation of the circle is $x^2 + y^2 + 2x + 4y + 3 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 1$ and $f = 2$.The center of the circle is $(-g, -f) = (-1, -2)$.The normal to a circle at any point always passes through the center of the circle.Thus,the normal passes through the points $(-1, -2)$ and $(-2, -3)$.The gradient (slope) $m$ of the normal is given by the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$.Substituting the coordinates: $m = \frac{-3 - (-2)}{-2 - (-1)} = \frac{-3 + 2}{-2 + 1} = \frac{-1}{-1} = 1$.Therefore,the gradient of the normal is $1$.

The gradient of the normal at the point $(-2, -3)$ on the circle $x^2 + y^2 + 2x + 4y + 3 = 0$ is

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