If the line x - y + 1 = 0 is a chord of the c | vedclass

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

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$\sqrt{14}$

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(D) The given circle equation is $x^2 + y^2 + 4y - 4 = 0$.
Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 0$,$f = 2$,and $c = -4$.
The center of the circle is $(-g, -f) = (0, -2)$.
The radius $r$ is $\sqrt{g^2 + f^2 - c} = \sqrt{0^2 + 2^2 - (-4)} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.
The perpendicular distance $p$ from the center $(0, -2)$ to the line $x - y + 1 = 0$ is calculated using the formula $p = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$.
$p = \frac{|1(0) - 1(-2) + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|0 + 2 + 1|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.
The length of the chord is given by $2\sqrt{r^2 - p^2}$.
Length $= 2\sqrt{(2\sqrt{2})^2 - (\frac{3}{\sqrt{2}})^2} = 2\sqrt{8 - \frac{9}{2}} = 2\sqrt{\frac{16 - 9}{2}} = 2\sqrt{\frac{7}{2}} = \sqrt{4 \cdot \frac{7}{2}} = \sqrt{14}$.

If the line $x - y + 1 = 0$ is a chord of the circle $x^2 + y^2 + 4y - 4 = 0$,find the length of this chord.

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