The normal drawn at (1,1) to the circle x^2+y | vedclass

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

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$4x+y=5$

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(B) The equation of the circle is $x^2+y^2-4x+6y-4=0$. Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-2$ and $f=3$. The center of the circle is $(-g, -f) = (2, -3)$. The normal to a circle at any point $(x_1, y_1)$ always passes through the center of the circle. Thus,the normal is the line passing through $(1, 1)$ and $(2, -3)$. The slope of this line is $m = \frac{-3-1}{2-1} = \frac{-4}{1} = -4$. The equation of the line is $(y-1) = -4(x-1)$. $y-1 = -4x+4$. $4x+y = 5$.

The normal drawn at $(1,1)$ to the circle $x^2+y^2-4x+6y-4=0$ is

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