The circle x^2 + y^2 = 4x + 8y + 5 intersects | vedclass

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(A) The given equation of the circle is $x^2 + y^2 - 4x - 8y - 5 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g =

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$ - 35 < m < 15$

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

The circle $x^2 + y^2 = 4x + 8y + 5$ intersects the line $3x - 4y = m$ at two distinct points if:

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