The slope of the normal to the circle x^2+y^2 | vedclass

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(D) The equation of the circle is $x^2+y^2+2gx+2fy+c=0$. The center of the circle is $C(-g, -f)$. The normal to a circle at any point $P(x_1, y_1)$ al

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$\frac{y_1+f}{x_1+g}$

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(D) The equation of the circle is $x^2+y^2+2gx+2fy+c=0$. The center of the circle is $C(-g, -f)$. The normal to a circle at any point $P(x_1, y_1)$ always passes through the center $C(-g, -f)$. The slope of the normal is the slope of the line segment $CP$. The slope $m$ is given by $\frac{y_2-y_1}{x_2-x_1} = \frac{y_1 - (-f)}{x_1 - (-g)} = \frac{y_1+f}{x_1+g}$. Thus,the slope of the normal is $\frac{y_1+f}{x_1+g}$.

The slope of the normal to the circle $x^2+y^2+2gx+2fy+c=0$ at $(x_1, y_1)$ is

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