The equation of the diameter of the circle x^ | vedclass

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(C) The given circle is $x^2 + y^2 + 2x - 4y - 11 = 0$. The center of this circle is $(-g, -f) = (-1, 2)$.The diameter that bisects a chord is perpend

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$x + 2y - 3 = 0$

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(C) The given circle is $x^2 + y^2 + 2x - 4y - 11 = 0$. The center of this circle is $(-g, -f) = (-1, 2)$.The diameter that bisects a chord is perpendicular to that chord.The given line is $2x - y + 3 = 0$,which has a slope of $m_1 = 2$.The slope of the diameter perpendicular to this line is $m_2 = -\frac{1}{m_1} = -\frac{1}{2}$.The equation of the line passing through the center $(-1, 2)$ with slope $-\frac{1}{2}$ is given by $y - 2 = -\frac{1}{2}(x + 1)$.Multiplying by $2$,we get $2y - 4 = -x - 1$.Rearranging the terms,we get $x + 2y - 3 = 0$.

The equation of the diameter of the circle $x^2 + y^2 + 2x - 4y - 11 = 0$ which bisects the chords intercepted on the line $2x - y + 3 = 0$ is

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