For the circle x^2 + y^2 - 2x + 4y - 4 = 0,wh | vedclass

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

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(A) The equation of the circle is $x^2 + y^2 - 2x + 4y - 4 = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$,$f = 2$,and $c = -4$. The center of the circle is $(-g, -f) = (1, -2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + 2^2 - (-4)} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$. To determine the nature of the line $2x - y - 1 = 0$ with respect to the circle,we calculate the perpendicular distance $d$ from the center $(1, -2)$ to the line: $d = \frac{|2(1) - (-2) - 1|}{\sqrt{2^2 + (-1)^2}} = \frac{|2 + 2 - 1|}{\sqrt{4 + 1}} = \frac{3}{\sqrt{5}}$. Since $d Therefore,the line is a chord of the circle.

For the circle $x^2 + y^2 - 2x + 4y - 4 = 0$,what is the line $2x - y - 1 = 0$?

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