If the line 3x - 4y = lambda touches the circ | vedclass

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

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$-35, 15$

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(B) The given circle is $x^2 + y^2 - 4x - 8y - 5 = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -4$,and $c = -5$.The center of the circle is $(-g, -f) = (2, 4)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{2^2 + 4^2 - (-5)} = \sqrt{4 + 16 + 5} = \sqrt{25} = 5$.Since the line $3x - 4y - \lambda = 0$ touches the circle,the perpendicular distance from the center $(2, 4)$ to the line must be equal to the radius $r$.Distance $d = \frac{|3(2) - 4(4) - \lambda|}{\sqrt{3^2 + (-4)^2}} = 5$.$\frac{|6 - 16 - \lambda|}{\sqrt{9 + 16}} = 5$.$\frac{|-10 - \lambda|}{5} = 5$.$|-10 - \lambda| = 25$.This implies $-10 - \lambda = 25$ or $-10 - \lambda = -25$.If $-10 - \lambda = 25$,then $\lambda = -35$.If $-10 - \lambda = -25$,then $\lambda = 15$.Thus,$\lambda = -35, 15$.

If the line $3x - 4y = \lambda$ touches the circle $x^2 + y^2 - 4x - 8y - 5 = 0$,then $\lambda$ is equal to

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