If the circle x^2+y^2-4x-8y-5=0 intersects th | vedclass

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(D) The equation of the circle is $x^2+y^2-4x-8y-5=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=-4, c=-5$. The center of the circle i

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$49$

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(D) The equation of the circle is $x^2+y^2-4x-8y-5=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=-4, c=-5$. The center of the circle is $(2, 4)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-2)^2+(-4)^2-(-5)} = \sqrt{4+16+5} = \sqrt{25} = 5$. For the line $3x-4y-m=0$ to intersect the circle at two distinct points,the perpendicular distance $d$ from the center $(2, 4)$ to the line must be less than the radius $r$. $d = \frac{|3(2)-4(4)-m|}{\sqrt{3^2+(-4)^2}} = \frac{|6-16-m|}{\sqrt{9+16}} = \frac{|-10-m|}{5} = \frac{|m+10|}{5}$. Setting $d This inequality is equivalent to $-25 Subtracting $10$ from all parts,we get $-35 The integers $m$ satisfying this are $\{-34, -33, \dots, 14\}$. The number of such integers is $14 - (-34) + 1 = 14 + 34 + 1 = 49$.

If the circle $x^2+y^2-4x-8y-5=0$ intersects the line $3x-4y-m=0$ in two distinct points,then the number of integral values of '$m$' is

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