The circle x^2 + y^2 - 4x - 8y - 5 = 0 will i | vedclass

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(C) The given circle is $x^2 + y^2 - 4x - 8y - 5 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -4$,and $c = -5$. The centr

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$ - 35 < m < 15$

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(C) The given circle is $x^2 + y^2 - 4x - 8y - 5 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -4$,and $c = -5$. The centre of the circle is $(-g, -f) = (2, 4)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 16 + 5} = \sqrt{25} = 5$. For the line $3x - 4y - m = 0$ to intersect the circle at two distinct points,the perpendicular distance from the centre $(2, 4)$ to the line must be less than the radius $r$. Distance $d = \frac{|3(2) - 4(4) - m|}{\sqrt{3^2 + (-4)^2}} = \frac{|6 - 16 - m|}{\sqrt{9 + 16}} = \frac{|-10 - m|}{5}$. We require $d $|-10 - m| $-25 Multiplying by $-1$,we get $25 > 10 + m > -25$. Subtracting $10$ from all parts,we get $15 > m > -35$,which is $-35 < m < 15$.

The circle $x^2 + y^2 - 4x - 8y - 5 = 0$ will intersect the line $3x - 4y = m$ in two distinct points,if:

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