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

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(B) The given equation of the circle is $x^2+y^2+6x+4y-3=0$. Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=3$ and $f=2$. The ce

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$y+2=0$

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(B) The given equation of the circle is $x^2+y^2+6x+4y-3=0$. Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=3$ and $f=2$. The center of the circle is $(-g, -f) = (-3, -2)$. The normal to a circle at any point always passes through the center of the circle. We need to find the equation of the line passing through the center $(-3, -2)$ and the given point $(1, -2)$. Since the $y$-coordinates of both points are equal to $-2$,the line is a horizontal line given by $y = -2$. Thus,the equation of the normal is $y+2=0$.

The equation of the normal to the circle $x^2+y^2+6x+4y-3=0$ at the point $(1, -2)$ is:

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