The equation of the chord of the circle x^2+y | vedclass

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(B) The equation of the circle is $x^2+y^2-4x-10y+25=0$.Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get the center $C = (-g, -f)

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$x+3y=7$

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(B) The equation of the circle is $x^2+y^2-4x-10y+25=0$.Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get the center $C = (-g, -f) = (2, 5)$.Let $M(1, 2)$ be the midpoint of the chord $AB$.The chord is perpendicular to the radius $CM$ at the point $M$.The slope of $CM$ is $m_{CM} = \frac{2-5}{1-2} = \frac{-3}{-1} = 3$.Since the chord $AB$ is perpendicular to $CM$,the slope of the chord $AB$ is $m_{AB} = -\frac{1}{m_{CM}} = -\frac{1}{3}$.The equation of the chord passing through $M(1, 2)$ with slope $-\frac{1}{3}$ is given by:$y - 2 = -\frac{1}{3}(x - 1)$$3(y - 2) = -(x - 1)$$3y - 6 = -x + 1$$x + 3y = 7$.

The equation of the chord of the circle $x^2+y^2-4x-10y+25=0$ having its midpoint at $(1,2)$ is

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