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

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(A) The given equation of the circle is $x^2+y^2-2x+4y-5=0$. Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$, we get $g=-1$ and $f=2$. The

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$y=3x-5$

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(A) The given equation of the circle is $x^2+y^2-2x+4y-5=0$. 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 $C$ is $(-g, -f) = (1, -2)$. The normal to a circle at any point on its circumference always passes through the center of the circle. Therefore, the normal is the line passing through the center $C(1, -2)$ and the given point $A(2, 1)$. The slope $m$ of the line passing through $(1, -2)$ and $(2, 1)$ is given by $m = \frac{1 - (-2)}{2 - 1} = \frac{3}{1} = 3$. The equation of the line passing through $(2, 1)$ with slope $m=3$ is $(y - 1) = 3(x - 2)$. Simplifying this, we get $y - 1 = 3x - 6$, which results in $y = 3x - 5$.

The equation of the normal to the circle $x^2+y^2-2x+4y-5=0$ at the point $(2,1)$ is:

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