If (3,-1) is one end of a diameter of the cir | vedclass

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

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$2x+y+5=0$

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(B) The given equation of the circle is $x^2+y^2-2x+4y=0$. 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 $C(-g, -f) = (1, -2)$. Let $A(3, -1)$ be one end of the diameter and $B(x_1, y_1)$ be the other end. Since the center $C$ is the midpoint of the diameter $AB$,we have $(1, -2) = (\frac{3+x_1}{2}, \frac{-1+y_1}{2})$. Solving for $B$,we get $3+x_1=2 \implies x_1=-1$ and $-1+y_1=-4 \implies y_1=-3$. So,the other end of the diameter is $B(-1, -3)$. The tangent at $B(-1, -3)$ is perpendicular to the radius $CB$. The slope of the radius $CB$ is $m_{CB} = \frac{-3-(-2)}{-1-1} = \frac{-1}{-2} = \frac{1}{2}$. The slope of the tangent at $B$ is $m = -\frac{1}{m_{CB}} = -2$. The equation of the tangent at $B(-1, -3)$ is $y - (-3) = -2(x - (-1))$. $y+3 = -2x-2$,which simplifies to $2x+y+5=0$.

If $(3,-1)$ is one end of a diameter of the circle $x^2+y^2-2x+4y=0$,then the equation of the tangent at the other end of that diameter is

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